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LOCUS 1 Straight Lines NOTES CONCEPT NOTES 01. Introduction 02. Straight Lines 03. Pair of Straight Lines Maths / Straight Lines LOCUS 2 Straight Lines Section - 1 INTRODUCTION Co-ordinate geometry is a marriage of pure geometry and algebra, and is indispensable in all branches of science today. Many of you must be pretty familiar with the general outline of this subject. We will restrict ourselves to 2-dimensional (plane) co-ordinate geometry in the following pages. Later on, we will also get the chance of studying 3-dimensional co-ordinate geometry. The basic idea in co-ordinate geometry, as has been mentioned earlier, is to study the properties of geometrical figures such as straight lines, circles, parabolas etc through the use of numbers. The core concept is that on a 2-dimensional (Euclidean) plane, any point can be represented by a pair of real numbers, using two non-parallel straight lines. The point where these two non-parallel reference lines meet is termed the origin of the reference axis. By convention, one axis is called the x-axis and one the y-axis. Any point on the plane can now be determined in reference to this reference axes as described in the figure below: y-axis The point P lies x units along the P x-axis and y units along the y-axis, and therefore, P can be represented as (x, y). Note that x, y ε R x is called the abscissa of P while y is y the ordinate of P. x and y together are called the co-ordinates of P. (origin) O x x-axis Fig - 1 Conversely, given the co-ordinates x and y of a point P, we can easily determine its location by moving x units along the x-axis and then y-units parallel to the y-axis. Notice that as long as the two axes are non-parallel, the entire plane is representable using these two axes as reference. These two axes in general can be at any non-zero angle to each other. Maths / Straight Lines LOCUS 3 However, it is almost always the case (out of convenience) that the two axis are taken at right angles to each other. Such axes are called Rectangular Axes. We will always be using Rectangular Axes in our discussion from now onwards. With this introduction, we start with the most elementary of geometric figures: line segments and lines (and other geometric figures obtainable from these elementary ones, like polygons). Section - 2 STRAIGHT LINES For a good command over co-ordinate geometry, a lot of results will be required to be memorised since they are encountered so often. For this purpose, each new theorem or result or property that we will encounter in the following pages is discussed in a separate article for ease of reference later. Art 1 : Distance formula One of the most basic expressions in co-ordinate geometry is that of the distance between two arbitrary points P ( x1 , y1 ) and P2 ( x2 , y2 ) . To obtain the required distance in terms of the co-ordinates of these 1 points, the Pythagoras theorem is employed as described in the figure below: y P2(x2 ,y2 ) d Note that: (i) OC = x1 ; OB = x2 P1 ⇒ BC = AP1 = x2 – x1 A (x1 ,y1) (ii) CP1 = y1 ; BP2 = y2 ⇒ AP2 = y2 – y1 x O C B Fig - 2 As explained in the figure, the distances AP1 and AP2 have been obtained. Thus, by the Pythagoras theorem, the distance d is AP 2 + AP22 or 1 d= ( x2 − x1 ) + ( y2 − y1 ) 2 2 As a direct consequence of this formula, the distance of an arbitrary point P(x, y) from the origin is x 2 + y 2 . As an elementary exercise, assume four points anywhere on the co-ordinate plane randomly, and use the distance formula to calculate the distance between each pair of points. Art 2 : Section formula Suppose that we are given two fixed points in the co-ordinate plane, say A ( x1 , y1 ) and B ( x2 , y2 ) . We need to find the co-ordinates of the point C which divides the line segment AB in the ratio m : n. Observe Maths / Straight Lines LOCUS 4 that two such points will exist. Name them C1 and C2. One of them will divide the line segment AB internally in the ratio m : n while the other will divide AB in the same ratio externally, as shown in the figure below: y y C2(xe , ye ) B(x2 , y2) B(x2 , y2 ) C1(xi , yi ) A(x1 , y1 ) A(x1 , y1 ) x x Internal division External division C1 divides AB internally in C2 divides AB externally in the ratio m : n. Thus, the ratio m : n. Thus, AC1 m AC2 m = = C1B n C2 B n Fig - 3 Let us find the co-ordinates of C1 using the help of the more detailed figure of internal division below: y (0, y2 ) B(x2 , y2 ) Note that AEC1 and ADB are C1(xi , yi ) similar. Thus, (0, yi ) AE EC1 AC1 = = AD DB AB A(x1 , y1) (0, y) D It is given that E AC1 : C1 B = m : n. Thus, AC1 m = AB m+n x O (x1,0) (xi ,0) (x2 ,0) Fig - 4 As described in the figure above, AE EC1 m = = AD DB m + n xi − x1 yi − y1 m ⇒ = = x2 − x1 y2 − y1 m + n mx2 + nx1 my2 + ny1 ⇒ , yi = xi = m+n m+n Thus, the co-ordinates of the point C1 which divides AB internally in the ratio m : n are Internal mx2 + nx1 my2 + ny1 , m+n m+n Division m:n Maths / Straight Lines LOCUS 5 Using an analogous approach, we can obtain the co-ordinates of the point C2 which divides AB externally in the ratio m : n External mx2 − nx1 my2 − ny1 , m−n m−n Division m:n A particular case of internal division is finding the co-ordinates of the mid-point of AB. Since the mid-point of AB divides the segment AB in the ratio 1 : 1, the co-ordinates of the mid-point will be Mid − point of AB where x1 + x2 y1 + y2 , A ≡( x1 , y1 ) and B ≡ ( x2 , y2 ) 2 2 Let us put these rudimentary results to use. Example – 1 Find the co-ordinates of the centroid of a triangle with the vertices A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ). Solution: To determine the centroid, we will borrow a result from plane geometry that you might remember from high school: the centroid divides any median in the ratio 2 : 1. A(x1, y1) AD is a median of ABC G divides AD in the ratio 2 : 1, i.e, AG : GD = 2 : 1 G D C(x3 , y3 ) B(x2 , y2 ) Fig - 5 x + x y + y3 The co-ordinates of D, the mid-point of BC, are 2 3 , 2 . Since AG : GD = 2 :1, 2 2 we have the co-ordinates of G by the section formula as x +x y + y3 2 2 3 + 1.x1 2 2 + 1. y1 G≡ 2 , 2 2 +1 2 +1 x1 + x2 + x3 y1 + y2 + y3 ≡ , 3 3 v The expression for the centroid confirms the obvious fact that the co-ordinates of the centroid are ‘symmetric’ with respect to the co-ordinates of the three vertices of the triangle. Maths / Straight Lines LOCUS 6 Example – 2 G is the centroid of triangle ABC. If O is any other point in the plane, prove that OA2 + OB 2 + OC 2 = GA2 + GB 2 + GC 2 + 3 GO 2 . Solution: There’s no loss of generality in taking O as the origin of our reference axis since even if we are given O to be a non-origin point, we can always translate the axes (keeping the triangle ABC unchanged) so that its origin coincides with O. Note that this operation will have no effect on the lengths OA, OB, OC , OG , GA, GB, GC etc. However, the expressions for distances will become much more simplified (In co-ordinate geometry, you will be required to follow such steps often, so that the expressions you are to deal with can be kept as simple as possible). Now, we assume some co-ordinates for A, B and C as shown in the figure below: y C (x3 , y3 ) B (x2 , y2 ) As discussed earlier, the co-ordinates of G (x, y) are G (x, y) x1 + x2 + x3 y1 + y2 + y3 , 3 3 x O A(x1 , y1 ) Fig - 6 We have, OA2 + OB 2 + OC 2 = x12 + y12 + x2 + y2 + x3 + y3 2 2 2 2 ... (1) while GA2 + GB 2 + GC 2 + 3GO 2 = ( x − x1 ) + ( y − y 1 ) + ( x − x2 ) + ( y − y2 ) 2 2 2 2 + ( x − x3 ) + ( y − y 3 ) + 3 ( x 2 + y 2 ) 2 2 = x12 + x2 + x3 + y12 + y2 + y3 2 2 2 2 ... (2) Comparing (1) and (2), we see that the two expressions are indeed equal v ____________________________________________________________________________________ Art - 3 Area of a triangle Suppose we are given three points in the co-ordinate plane : A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ). We intend to find the area of ∆ ABC in terms of the given co-ordinates. How to evaluate this area is described in the figure below: Maths / Straight Lines LOCUS 7 y B (x2 , y2 ) Note that area ( ABC) = area (trapezium APQB) + ) 1 ,y 1 area (trapezium BQRC) x A( C (x3 , y3) – area (trapezium APRC) x O P Q R Fig - 7 Observe how the area of ∆ ABC has been written in terms of the area of three trapeziums. 1 From plane geometry, the area of a trapezium is × (sum of bases ) × height. Thus, 2 1 area ( trap. APQB ) = × ( AP + BQ ) × PQ 2 1 = ( y1 + y2 )( x2 − x1 ) ... (1) 2 Similarly, 1 area ( trap. BQRC ) = × ( BQ + CR ) × QR 2 1 = ( y2 + y3 )( x3 − x2 ) ... (2) 2 1 area ( trap. APRC ) = × ( AP + CR ) × PR 2 1 ( y1 + y3 )( x3 − x1 ) = ... (3) 2 From (1), (2) and (3), we have, upon simplification, 1 area ( ∆ ABC ) = ( x2 y1 − x1 y2 + x3 y2 − x2 y3 + x1 y3 − x3 y1 ) 2 {x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )} 1 = 2 We used the mod sign in the last expression because area is by definition positive. We can express the area obtained in determinant form very concisely: x1 y1 1 1 ∆ = area ( ∆ ABC ) = x2 y2 1 2 x3 y3 1 Maths / Straight Lines LOCUS 8 As a consequence of this result, we can now easily find the area of an arbitrary quadrilateral ABCD as describe in the figure below: y B(x2 , y2 ) C(x3 , y3 ) area (quad ABCD) = area ( ABC) + area ( ACD) x1 y1 1 x1 y1 1 1 D(x4, y4) = x2 y2 1 + x3 y3 1 2 x3 y3 1 x4 y4 1 A(x1 , y1 ) In other words, we evaluate the areas of the two triangles x separately and add them to O get the area of the quadrilateral Fig - 8 We can generalise this method easily to find the area of any polygon as a sum of the areas of the constituent triangles. v Example – 3 a a a Find the area of the triangle, the co-ordinates of whose vertices are ap, , aq, and ar , . p q r Solution: Using the result obtained is Art - 3, we have, a ap 1 p 1 a ∆= aq 1 2 q a ar 1 r 1 a a a a a a = ap − + aq − + ar − 2 q r r p p q a2 p (r − q ) q ( p − r ) r (q − p ) = + + 2 qr pr pq a 2 p 2 (q − r ) + q 2 (r − p ) + r 2 ( p − q ) = 2 pqr v Maths / Straight Lines LOCUS 9 Example – 4 Assume two fixed points in the co-ordinate plane: A ( a, 0 ) and B ( −a, 0 ) . A variable point C(x, y) moves in the plane in such a way that CA + CB is a constant k. Use the distance formula to evaluate the condition that the co-ordinates of C must satisfy. Solution: We have, CA = (x − a) + y2 2 CB = (x + a) + y2 2 and From the constraint specified in the question, we have CA + CB = k ⇒ CA2 + CB 2 + 2 CA ⋅ CB = k 2 ⇒ (x − a) 2 + y2 + ( x + a ) + y2 + 2 2 (( x − a ) + y ) (( x + a ) + y ) = k 2 2 2 2 2 (x − a2 ) + y 4 + 2 y 2 ( x2 + a2 ) = k 2 − 2 ( x2 + y 2 + a2 ) 2 ⇒ 2 2 ⇒ 2 x 4 + y 4 + a 4 + 2 x 2 y 2 − 2a 2 x 2 + 2a 2 y 2 = k 2 − 2 ( x 2 + y 2 + a 2 ) Squaring both sides and cancelling out the common terms on both sides, we obtain −8a 2 x 2 = k 4 + 8a 2 x 2 − 4 k 2 ( x 2 + y 2 + a 2 ) ⇒ 4k 2 x 2 − 16a 2 x 2 + 4k 2 y 2 = k 4 − 4k 2 a 2 ⇒ 4 x 2 ( k 2 − 4a 2 ) + 4 k 2 y 2 = k 2 ( k 2 − 4a 2 ) This is the relation that the co-ordinates of the variable point C (x, y) must satisfy. All the pairs (x, y) which satisfy this equation, when plotted on the co-ordinate plane, will trace out the path on which the variable point C is constrained to move. In other words, this equation specifies the locus of the point C. ____________________________________________________________________________________ Art - 4 Equation ( s ) representing a straight line The last three articles dealt with the preliminaries of co-ordinate geometry and certain elementary formulae which find widespread use. With this article, we start the discussion of the geometry of straight lines in detail. On the co-ordinate plane, the simplest case for a straight line would be one in which the line is parallel to one of the co-ordinate axes. Maths / Straight Lines LOCUS 10 y y y0 x x0 x O O A line parallel to the x-axis. A line parallel to the y-axis. Any point on this line has a constant Any point on this line has a constant y-co-ordinate equal to y0. x-co-ordinate equal to x0. Thus, we can say that the equation Thus, we can say that the equation of this line is y = y0 of this line is x = x0 (There's no constraint on the (There's no constraint on the x-co-ordinate of this line) y-co-ordinate of this line) Fig - 9 As described in the figure above, the equation of such a line is y = y0 or x = x0 accordingly as the line is parallel to the x-axis or the y-axis respectively. These are special cases of lines; we want to find the equation of any arbitrary line in general. Visualise any such line in your mind. To completely specify such a line, you would need two quantities: the inclination of the line (or its slope or the angle it makes with say, the x-axis) and the placement of the line (i.e. where the line passes through with reference to the axes: we can specify the placement of the line by specifying the point on the y-axis through which the line passes, or in other words, by specifying the y-intercept.) y We need this length c which tells us the placement of the line θ x We need this angle which tells us the inclination of the line Fig - 10 It should be obvious to you that any line can be determined uniquely using these two parameters. We now find out the equation of this straight line, assuming that we know θ and c. In other words, we intend to find out the relation that the co-ordinates (x, y) of any arbitrary point on the line must satisfy. The determination of this equation is straightforward: Maths / Straight Lines LOCUS 11 y P We assume an arbitrary (x, y) point P(x, y) on the line and try to relate x and y θ to the known quantities A(0, c) B θ and c. The relation we require is obtainable from the fact that in APB, θ x O PB C tan θ= AB Fig - 11 As described in the figure above, we have in ∆ APB , PB tan θ = AB tan θ is a measure of the inclination of the line (its steepness). tan θ is therefore termed the slope of the line and is denoted by m. Thus, m = tan θ . Also, notice that PB = ( y − c ) and AB = x. Therefore, y−c m= x ⇒ y = mx + c : Slope - intercept form ! ! Slope y − intercept This is the general equation of a straight line involving its slope and its y-intercept. This form of the equation of the line is therefore termed the slope-intercept form. Notice that if the line passes through the origin, its equation would reduce to y = mx. As you might have guessed by now, this is not the only form to represent a straight line. This form uses the slope and the intercept of the line. Lets discuss another form. Notice that to uniquely determine any straight line, we either need the slope of the line and a point through which this line passes, or we need at least two points through which that line passes. Thus for example, a line can also be uniquely determined if we are given the two points where this line intersects the x-axis and the y-axis. y The straight line L can be uniquely determined if we know a and b, i.e if we know the x-intercept and the y - intercept b φ θ x O a L Fig - 12 Maths / Straight Lines LOCUS 12 b b Notice that tan φ = so that the slope of the line is m = tan θ = tan ( π − φ ) = − tan φ = − . Also, the a a y-intercept is b. Thus, using the slope intercept form obtained earlier, the equation of the line L is b y=− x+b a ⇒ bx + ay = ab x y : Intercept form ⇒ + =1 a b ! " x − intercept y − intercept Thus, if we know the x and y intercepts, we can directly use this form to write the equation of the line. Lets consider a third form to represent a line. From the figure below, observe carefully that to uniquely determine a line, we could also specify the length of the perpendicular dropped from the origin to that line and the orientation (inclination) of that perpendicular: y The straight line L can be uniquely determined if we know p and α p α y L Fig - 13 To determine the equation of this line, assume any point P on the line with the co-ordinate (x, y). Observe the geometry described in the figure below carefully: y Observe that: OR = OQ cos α = x cos α A and RA = SP = PQ sin α = y sin α P(x, y) p R S α α x O Q B Fig - 14 Maths / Straight Lines LOCUS 13 From the figure, note that OR + RA = OA = p ⇒ x cos α + y sin α = p : Normal form ! " inclination of length of perpendicular perpendicular Thus, we now know of three forms in which the equation of an arbitrary straight line can be written. From those three forms, you might be able to deduce that the most general form for the equation of an arbitrary straight line is Ax + By + C = 0 . Let us try to prove this assertion, that is, let us try to show that Ax + By + C = 0 represents the equation of a straight line. For this purpose, it will suffice to show that if we take any three arbitrary points ( x1 , y1 ) , ( x2 , y2 ) and ( x3 , y3 ) on the curve Ax + By + C = 0, these three points will turn out to be collinear. Equivalently, the area of the triangle with the vertices as these three points will turn out to be zero. Since all the three points satisfy the equation Ax + By + C = 0, we have Ax1 + By1 + C = 0 Ax2 + By2 + C = 0 Ax3 + By3 + C = 0 We can eliminate A, B and C from these three equations simultaneously to obtain a relation involving only the co-ordinates of the three points. A basic knowledge of elimination in determinant form will tell you that the relation we’ll get after elimination is x1 y1 1 x2 y2 1 = 0 x3 y3 1 which means that the area of the triangle formed by these three points as vertices is zero! Hence, the assertion is true. With this discussion in mind, you should be able to write the equation for any arbitrary straight line. We will encounter the use of all these forms in the coming examples. Before concluding this article, do this as a simple exercise based on the discussion we've already done: (a) Show that the equation of the straight line with slope m and passing through the fixed point (x1, y1) is y − y1 = m ( x − x1 ) (b) Show that the equation of the straight line passing through the two fixed points ( x1 , y1 ) and ( x2 , y2 ) is y − y1 y2 − y1 = x − x1 x2 − x1 Maths / Straight Lines LOCUS 14 The following table summarizes the various forms of the straight line that we've encountered. Known parameters about the line Equation Name of this form Slope m 1. y = mx + c Slope-intercept form y-intercept c x -intercept a x y 2. y -intercept b + =1 Intercept form a b Length of perpendicular 3. from origin to the line :p x cos α + y sin α = p Normal form Inclination of perpendicular :α Slope : m 4. Any point through y − y1 = m ( x − x1 ) Point-slope form which the line passes : ( x1 , y1 ) Any two points through : ( x1 , y1 ) y − y1 y2 − y1 5. = Two point form which the line passes : ( x2 , y2 ) x − x1 x2 − x1 Most general form : Ax + By + C = 0 where A, B, C ∈ # and at least one of A, B is non-zero Note that each of the five specific forms mentioned in the table above can be converted easily to the most general form of the equation of a line. You are urged to do this as an exercise. Also, the five forms are inter convertible among themselves in most cases too. For example, y = mx + c x y c can be written in intercept form as + = 1 so that the x-intercept of this line is a = − and the − (c / m ) c m y-intercept is b = c. You are urged to try out all the (possible) conversions from one form to another. You should now be able to understand that to determine a straight line uniquely, we must have two quantities given. Thus, two points could uniquely fix a line, or a point on the line and its slope could do so too, and so on. Notice that the general equation of the line also in fact contains only two arbitrary constants: Ax + By + C = 0 A B ⇒ x + y +1 = 0 C C contains only two ⇒ Px + Qy + 1 = 0 arbitrary constants Art 5 Point of intersection ; Angle of intersection We are given two lines L1 and L2, and we are required to find the point at which they intersect (if they are non-parallel) and the angle at which they are inclined to one another, i.e., the angle of intersection. Evaluating the point of intersection is a simple matter of solving two simultaneous linear equations. Let the Maths / Straight Lines LOCUS 15 equations of the two lines be L1 ≡ a1 x + b1 y + c1 = 0 and L2 ≡ a2 x + b2 y + c2 = 0 (written in the most general form). Now, let the point of intersection be (x1, y1). Thus, a1 x1 + b1 y1 + c1 = 0 a2 x1 + b2 y1 + c2 = 0 This system can be solved to get x1 y1 1 = = b1c2 − b2 c1 c1a2 − c2 a1 a1b2 − a2b1 From this relation we obtain the point of intersection ( x1 , y1 ) as b1c2 − b2 c1 c1a2 − c2 c1 , : Point of intersection a1b2 − a2b1 a1b2 − a2b1 To obtain the angle of intersection between these two lines, consider the figure below: y L2 = a2 x + b2 y + c2 = 0 Note that θ = θ2− θ1 L1 = a1 x + b1 y + c1 = 0 θ θ1 θ2 x Fig - 15 The equation of the two lines in slope-intercept form are a c c a y1 = − 1 x + 1 = m1 x + 1 where m1 = − 1 b1 b1 b1 b1 a c c a and y2 = − 2 x + 2 = m2 x + 2 where m2 = − 2 b2 b2 b2 b2 Note in Fig - 15 that θ = θ2 − θ1 and thus tan θ = tan ( θ2 − θ1 ) tan θ2 − tan θ1 = 1 + tan θ1 tan θ2 m2 − m1 = ... (1) 1 + m1m2 Conventionally, we would be interested only in the acute angle between the two lines and thus we have to m2 − m1 have tan θ as a positive quantity. So in (1) above, if the expression 1 + m m turns out to be negative, this 1 2 would be the tangent of the obtuse angle between the two lines; thus, to get the acute angle between the two lines, we use the magnitude of this expression. Maths / Straight Lines LOCUS 16 Therefore, the acute angle θ between the two lines is m2 − m1 θ = tan −1 : Acute angle between the two lines 1 + m1 m2 From this relation, we can easily deduce the conditions on m1 and m2 such that the two lines L1 and L2 are parallel or perpendicular. If the lines are parallel, θ = 0 so that m1 = m2 which is intuitively obvious since parallel lines must have the π same slope. For the two lines to be perpendicular, θ = so that cot θ = 0; this can happen if 1 + m1m2 = 0 2 or m1m2 = −1. Thus, m1 = m2 : for parallel lines and m1 m2 = −1 : for perpendicular lines If the lines L1 and L2 are given in the general form given in the general form ax + by + c = 0, the slope of a this line is m = − so that the condition for L1 and L2 to be parallel becomes − a1 = − a2 or a1b2 = a2b1 b b1 b2 aa and the condition for L1 and L2 to be perpendicular becomes 1 2 = −1 or a1a2 + b1b2 = 0 . b1b2 For example, the line L1 ≡ x + y = 1 is perpendicular to the line L2 ≡ x − y = 1 because the slope of L1 is –1 while the slope of L2 is 1. y L2 = x - y = 1 x L1 = x + y = 1 Fig - 16 As another example, the line L1 ≡ x − 2 y + 1 = 0 is parallel to the line L2 ≡ x − 2 y − 3 = 0 because the 1 slope of both the lines is m = . 2 Maths / Straight Lines LOCUS 17 Example – 5 Find the equation to the straight line which passes through (3, –2) and is inclined at 60º to the line 3 x + y = 1. Solution: Observe carefully that there will be two such lines. Denote the two lines by L1 and L2 L2 60º L1 (3, -2) 60º 3x+y = 1 Fig - 17 Let the slope of the line(s) we require be m. The slope of 3 x + y = 1 is m1 = − 3 Since we want the acute angle between the two lines to be 60º, we must have by Art - 5, m1 − m tan 60º = 1 + mm1 − 3−m ⇒ 3= 1− 3 m m+ 3 ⇒ =± 3 1− 3 m ⇒ m + 3 = 3 − 3m or m + 3 = 3m − 3 ⇒ m = 0 or m = 3 Since we get two values of m, this confirms our earlier assertion that two such lines will exist. We now have the slope. We also know that the lines pass through (3, –2). We can therefore use the point-slope form to write down the required equations: L1 ≡ y − ( −2 ) = 0 ( x − 3) ; L2 ≡ y − ( −2 ) = 3 ( x − 3) ⇒ L1 ≡ y + 2 = 0 and L2 ≡ y − 3 x + 2 + 3 3 = 0 Example – 6 Find the equation of the straight line which passes through the point ( a cos θ, a sin θ ) and is perpendicular to 3 3 the straight line x sec θ + y cosecθ = a . Maths / Straight Lines LOCUS 18 sec θ Solution: The slope of the given line is m1 = − = − tan θ cosecθ Therefore, the slope of the line we require will be given by m2 where 1 m2 = − m1 ⇒ m2 = cot θ We now know the slope of the line and we are also given a fixed point through which the line passes. We can therefore use the point-slope form to determine its equation: y − a sin 3 θ = cot θ ( x − a cos3 θ ) ⇒ x cos θ − y sin θ = a ( − sin 4 θ + cos 4 θ ) = a (cos 2 θ + sin 2 θ )(cos 2 θ − sin 2 θ ) = a cos 2θ Thus, the required equation is x cos θ − y sin θ = a cos 2θ ____________________________________________________________________________________ Art 6 Half-planes Any straight line divides the Euclidean plane into two half planes. In this article, we wish to determine the half-plane in which an arbitrary point lies with respect to a given line. Let the equation of the given line be ax + by + c = 0. Consider two points ( x1 , y1 ) and ( x2 , y 2 ) that lie in different half-planes with respect to this line: y ax + by + c = 0 (x2 , y2 ) P Half-plane B (x1 , y'1 ) Half-plane A Q (x1 , y1 ) (x2 , y'2 ) x Fig - 18 The point ( x1 , y1 ) lies in the lower half-plane while ( x2 , y2 ) lies in the upper half plane. We require a condition on these co-ordinates which must be satisfied if the points lie in opposite half-planes. In Fig - 18, we have dropped verticle line segments from ( x1 , y1 ) and ( x2 y2 ) to the given line, intersecting it in P and Q respectively. The co-ordinates of P and Q are ( x1 , y1' ) and ( x2 , y2 ) respectively where y1' ≠ y1 and y2' ≠ y2 . ' Since P, Q lie on the given line, their co-ordinates must satisfy the equation of the line. Thus, Maths / Straight Lines LOCUS 19 ax1 + by1' + c = 0 ⇒ y1' = − ( ax1 + c ) b ax2 + by2 + c = 0 ' ⇒ ' y2 =− ( ax2 + c ) and b Now, from Fig - 18 we have y1 < y1' and y2 > y2 ' ⇒ y1 < − ( ax1 + c ) y2 > − ( ax2 + c ) and b b ax1 + by1 + c ax2 + by2 + c ⇒ <0 and >0 b b ⇒ ax1 + by1 + c and ax2 + by2 + c are of opposite signs. This is the required condition. Translated into words, it says that for two points lying in opposite half-planes, their co-ordinates when substituted respectively into the equation of the line must give expressions of opposite signs. (For two points in the same half-plane, the signs would be the same). As a corollary, observe that a point (x1, y1) lies in the same half–plane or opposite half–plane in which the origin lies accordingly as ( ax1 + by1 + c ) and c are of the same sign or opposite signs respectively. Art 7 Length of perpendicular Suppose that we are given the equation of a line L and we are required to find the length of the perpendicular dropped from an arbitrary point P ( x1 , y1 ) on L. Suppose that the equation of L is in normal form, i.e, L ≡ x cos α + y sin α = p. y We are required to find PQ. Note that PQ = OR - OS R = OR - p To determine OR, we P (x1, y1) draw a line L' parallel to S L through P (x1, y1) Let OR = p1 p α Q x O L = x cos α + y sin α - p = 0 L' Fig - 19 Based on the discussion in the figure above, the equation of the line L' is x cos α + y sin α − p1 = 0. Since L1 passes through P, the co-ordinates of P must satisfy the equation of L1. Thus, x1 cos α + y1 sin α − p1 = 0 Thus, we get p1 as ( x1 cos α + y1 sin α ). The length of perpendicular PQ is now simply p1 − p = x1 cos α + y1 sin α − p . (Modulus sign is used since PQ is a length so it must be positive). PQ = x1 cos α + y1 sin α − p : Length of perpendicular Maths / Straight Lines LOCUS 20 Let us now assume the case where L is given in the general form, i.e. L ≡ ax + by + c = 0. We can easily adjust the equation of L so that c is negative. We do this so that we can convert L into the normal form: ax + by + c = 0 c<0 ⇒ ax + by = − c a b −c ⇒ 2 x+ 2 y = 2 a +b a +b a +b 2 2 2 ⇒ x cos α + y sin α = p ... (1) a b −c where cos α = , sin α = and p = a +b 2 2 a +b 2 2 a + b2 2 The equation in (1) is in the normal form; we can now use the result obtained in the preceding discussion to obtain the length of the perpendicular PQ: Modulus sign is PQ = x1 cos α + y1 sin α − p used since PQ must be +ve ax1 by1 c = + + a 2 + b2 a 2 + b2 a 2 + b2 ax1 + by1 + c PQ = : Length of perpendicular a 2 + b2 Example – 7 Find the distance between two parallel lines L1 and L2 given by L1 ≡ ax + by + c1 = 0 and L2 ≡ ax + by + c2 = 0 Solution: L1 Assume any point P on the line L1; we are P(x1, y1 ) required to find d d L2 Fig - 20 Since P lies on L1, we have ax1 + by1 + c1 = 0 ⇒ ax1 + by1 = −c1 ⇒ ax1 + by1 + c2 = c2 − c1 ... (1) Maths / Straight Lines LOCUS 21 By the previous article, the length of the perpendicular dropped from P upon the line L2 is ax1 + by1 + c2 d= a 2 + b2 c1 − c2 = (By (1) above) a 2 + b2 This is the required distance between the two lines Example – 8 If p and p' be the perpendiculars from the origin upon the lines L1 ≡ x sec θ + y cosec θ − a = 0 and L2 ≡ x cos θ − y sin θ − a cos 2θ = 0, show that 4 p 2 + p '2 = a 2 Solution: The length of the perpendicular dropped from (0, 0) to L1 is by Art - 7 a p= sec 2 θ + cosec 2θ = a sin θ cos θ ... (1) Similarly, p', the length of the perpendicular from (0, 0) to L2 is a cos 2θ p' = cos 2 θ + sin 2 θ = a cos 2θ ... (2) We now have from (1) and (2) 4 p 2 + p '2 = 4a 2 sin 2 θ + a 2 cos 2 2θ = a 2 sin 2 2θ + a 2 cos 2 2θ = a2 ____________________________________________________________________________________ Art 8 Concurrency Consider three different straight lines L1, L2 and L3: L1 ≡ a1 x + b1 y + c1 = 0 ... (1) L2 ≡ a2 x + b2 y + c2 = 0 ... (2) L3 ≡ a3 x + b3 y + c3 = 0 ... (3) We need to evaluate the constraint on the coefficients ai' s, bi' s and ci' s such that the three lines are concurrent. Let us first determine the point P of intersection of L1 and L2. By Art - 5, it will be b1c2 − b2 c1 c1a2 − c2 a1 P≡ , a1b2 − a2b1 a1b2 − a2b1 Maths / Straight Lines LOCUS 22 Thus three lines will be concurrent if L3 passes through P too, that is P satisfies the equation of L3. Thus, bc −b c c a −c a a3 1 2 2 1 + b3 1 2 2 1 + c3 = 0 a1b2 − a2b1 a1b2 − a2b1 ⇒ a3 (b1c2 − b2c1 ) + b3 ( c1a2 − c2 a1 ) + c3 ( a1b2 − a2b1 ) = 0 ⇒ a1 (b2 c3 − b3c2 ) + b1 ( c2 a3 − c3 a2 ) + c1 ( a2b3 − a3b2 ) = 0 This relation can be written compactly in determinant form as a1 b1 c1 a2 b2 c2 = 0 a3 b3 c3 This is the condition that must be satisfied for the three lines to be concurrent. For example, consider the three lines 2 x − 3 y + 5 = 0, 3 x + 4 y − 7 = 0 and 9 x − 5 y + 8 = 0 . These three lines are concurrent because the determinant of the coefficients is 0, i.e, 2 −3 5 3 4 −7 = 0 9 −5 8 Example – 9 Prove that the three lines L1, L2 and L3 whose equations have been mentioned in the preceeding discussion, are concurrent if we can find three constants λ1 , λ 2 and λ 3 such that λ1 L1 + λ 2 L2 + λ3 L3 = 0 Solution: Assume that L1 and L2 intersect at the point P whose co-ordinates are (x0, y0) P should satisfy the equations of both L1 and L2. L1 (at P ) ≡ a1 x0 + b1 y0 + c1 = 0 ... (1) L2 ( at P ) ≡ a2 x0 + b2 y0 + c2 = 0 ... (2) Now assume that we can find three non-zero constants λ1 , λ 2 and λ 3 such that λ1L1 + λ 2 L2 + λ 3 L3 = 0 . We will prove that due to this condition, L3 will definitely have to pass through P: λ1 L1 + λ 2 L2 + λ 3 L3 = 0 λ λ ⇒ L3 = − 1 L1 + − 2 L2 λ3 λ3 If we evaluate the value of L3 at P, we get λ λ L3 ( at P ) = − 1 × L1 (at P ) + − 2 × L2 (at P ) λ3 λ3 λ λ By (1) = − 1 ×0 + − 2 ×0 λ3 λ3 and ( 2 ) =0 Since the value of L3 is 0 at P, the line L3 must pass through P. Thus, L1, L2 and L3 are concurrent. Maths / Straight Lines LOCUS 23 Example – 10 Show that the medians of a triangle are concurrent. Solution: Let the triangle have the vertices A ( x1 , y1 ) , B ( x2 , y2 ) and C ( x3 , y3 ) as in the figure below: A(x1 , y1 ) Let D, E, F be the mid-points of BC, CA and AB respectively. E The co-ordinates of any mid point can easily be evaluated F by the section formula. For eg, C(x3 , y3 ) the co-ordinates of D are x2 + x3 y2 + y3 D , 2 2 B(x2 , y2 ) Fig - 21 From the two-point form of the equation of a line, we can write down the equations of AD, BE and CF. The equation L1 of the median AD is: y + y3 y1 − 2 y − y1 2 L1 ≡ = x − x1 x − x2 + x3 1 2 ⇒ L1 ≡ ( 2 y1 − ( y2 + y3 )) x − ( 2 x1 − ( x2 + x3 )) y = x1 ( 2 y1 − ( y2 + y3 )) − y1 ( 2 x1 − ( x2 + x3 )) By symmetry, we can write down the corresponding equations L2 and L3 of the medians BE and CF. Observe carefully that when we subsequently add the three equations L1, L2 and L3, their left hand sides sum to 0. Thus, we have found three constants 1, 1 and 1 such that 1 ⋅ L1 + 1 ⋅ L2 + 1 ⋅ L3 = 0 ⇒ L1 , L2 and L3 are concurrent ⇒ The medians of any triangle are concurrent. Example – 11 Show that the equation of any line passing through the intersection point P of two given lines whose equations are L1 and L2, can be expressed as L1 + λ L2 = 0, where λ is a real parameter. Solution: Let L1 ≡ a1 x + b1 y + c1 = 0 and L2 ≡ a2 x + b2 y + c2 = 0 Consider the equation L1 + λ L2 = 0 ... (1) ⇒ a1 x + b1 y + c1 + λ ( a2 x + b2 y + c2 ) = 0 ⇒ ( a1 + λa2 ) x + (b1 + λb2 ) y + (c1 + λc2 ) = 0 Maths / Straight Lines LOCUS 24 This is definitely the equation of a straight line because it is of the form ax + by + c = 0. Also, notice in addition that the intersection point P will satisfy this equation, because if we substitute the co-ordinates of the intersection point P in (1), both L1 and L2 vanish. Thus, L1 + λL2 = 0 is the equation of an arbitrary straight line that passes through the intersection point P of L1 and L2. (As we vary λ , the slope of this line will vary but it will always pass through P). The equation of any line passing through P can be L1 + λ L2= 0 written as L1 + λ L2 = 0 where λ is a real P parameter L1 = 0 L2 = 0 Fig - 22 This result is very beneficial in certain cases. We’ll see such cases in some subsequent examples Example – 12 Find the equations to the straight lines passing through (a) (3, 2) and the point of intersection of 2 x + 3 y = 1 and 3 x − 4 y = 6 x y x y (b) Origin and the point of intersection of + = 1 and + = 1 . a b b a Solution: (a) The equations of the two given lines in standard form are : L1 ≡ 2 x + 3 y − 1 = 0 L2 ≡ 3x − 4 y − 6 = 0 Any line passing through the intersection point of L1 and L2 is L1 + λ L2 = 0 ⇒ (2 x + 3 y − 1) + λ (3 x − 4 y − 6) = 0 ⇒ (2 + 3λ ) x + (3 − 4λ ) y − (1 + 6λ ) = 0 ... (1) We want this line to pass through (3, 2). Therefore (3, 2) must satisfy the equation of this line, i.e. (2 + 3λ )3 + (3 − 4λ )2 − (1 + 6λ ) = 0 ⇒ −5λ + 11 = 0 11 ⇒ λ= 5 Maths / Straight Lines LOCUS 25 11 We substitute λ = in (1) to get the required equation: 5 11 11 11 (2 + 2 ⋅ ) x + (3 − 4 ⋅ ) y − (1 + 6 ⋅ ) = 0 5 5 5 ⇒ 43 x − 29 y − 71 = 0 (b) We follow the same procedure as in part (a) L1 : bx + ay − ab = 0 L2 : ax + by − ab = 0 The equation of any line passing through the intersection point of L1 and L2 is L1 + λ L2 = 0 ⇒ (b + λ a ) x + (a + λ b ) y − ab (1 + λ ) = 0 ... (2) Since this line must pass through (0, 0), we substitute (0, 0) into (2) to get ab(1 + λ ) = 0 ⇒ λ = −1 We substitute λ = −1 into (2) to get the required equation : (b − a ) x + (a − b ) y = 0 ⇒ x− y =0 ____________________________________________________________________________________ Art 9 Angle Bisectors Consider two straight lines L1 and L2 with the equations L1 : a1 x + b1 y + c1 = 0 L2 : a2 x + b2 y + c2 = 0 We intend to find the angle bisector formed at the intersection point P of L1 and L2. Note that there will be two such angle bisectors y P We denote the two angle bisectors by A1 and A2 x A2 L2 A2 L1 Fig - 23 Maths / Straight Lines LOCUS 26 To write down the equations of the two angle bisectors, we first modify the equations of L1 and L2 so that c1 and c2 are say, both negative in sign. This can always be done. Why this is done will soon become clear. We first write down the equation of A1, the angle bisector of the angle in which the origin lies. By virtue of being an angle bisector, if any point P ( x′, y ′) lies on A1, the distance of P from L1 and L2 must be equal. Using the perpendicular distance formula of Art -7, we have a1 x′ + b1 y′ + c1 a2 x′ + b2 y′ + c2 ⇒ = a12 + b1 2 a22 + b22 a1 x′ + b1 y ′ + c1 a2 x′ + b2 y′ + c2 ⇒ =± ...(1) a1 + b1 2 2 a22 + b22 Which sign should we select, “+” or “–”, for the bisector of the angle containing the origin? Since P and origin lie on the same side of L1, a1 x′ + b1 y′ + c1 and c1 must be of the same sign by Art - 6. Similarly, a2 x′ + b2 y′ + c2 and c2 must be of the same sign. But since we have already arranged c1 and c2 to be of the same sign (both negative), we must have ( a1 x′ + b1 y′ + c1 ) and ( a2 x′ + b2 y′ + c2 ) also of the same sign. Thus, it follows from (1) that to write the equation of the angle bisector of the angle containing the origin, we must select the “+” sign since ( a1 x′ + b1 y′ + c1 ) and ( a2 x′ + b2 y′ + c2 ) are of the same sign. The “–” sign gives the angle bisector of the angle not containing the origin, i.e., the equation of A2. To summarize, we first arrange the equations of L1 and L2 so that c1 and c2 are both of the same sign. Subsequently, using the property of any angle bisector, we obtain a1 x + b1 y + c1 a2 x + b2 y + c2 Angle bisector of angle =+ : a1 + b1 2 2 a +b 2 2 contaning the origin 2 2 and a1 x + b1 y + c1 a2 x + b2 y + c2 Angle bisector of angle not =− : a1 + b1 2 2 a +b 2 2 contaning the origin 2 2 Example – 13 Find the angle bisector of the angle between the straight lines L1 : 3x − 4 y + 7 = 0 and L2 :12 x − 5 y − 8 = 0 which contains the origin. Maths / Straight Lines LOCUS 27 Solution: Following the discussion of the preceeding article, we first modify the equations L1 and L2 so that the constant terms in both the equations are of the same sign (say both positive): L1 : 3x − 4 y + 7 = 0 L2 : − 12 x + 5 y + 8 = 0 The angle bisector of the angle containing the origin is (3 x − 4 y + 7) (−12 x + 5 y + 8) =+ 32 + 4 2 12 2 + 52 ⇒ 99 x − 77 y + 51 = 0 Evaluating the other angle bisector is left to the reader as an exercise. Example – 14 Find the bisector of the angle between the lines x + 2 y − 11 = 0 and 3 x − 6 y − 5 = 0 which contains the point (1, –3). Solution: Again we first arrange the equations of the two lines such that constant terms are positive L1 : − x − 2 y + 11 = 0 L2 : − 3x + 6 y + 5 = 0 Note that L1 (at origin) > 0 and L1 (at(1, − 3)) > 0 ⇒ Origin and (1, –3) are on the same side of L1. L2 (at origin) > 0 and L2 (at(1, − 3)) < 0 ⇒ Origin and (1, –3) are on the opposite sides of L2. This means that the point (1, –3) does not lie in the same region as the origin, since (1, –3) must be on the opposite side of the origin with respect to L2.The example figure below will make this clear: L1 We see that (1, -3) lies in (1, -3 ) the angle not containing the origin Origin L2 Fig - 24 Maths / Straight Lines LOCUS 28 Thus, it is clear that (1, –3) lies in the angle not containing the origin. There is one point worth mentioning here. If suppose a point P (x, y) lies on the opposite side of the origin with respect to both L1 and L2, it would lie in the vertically opposite angle to the angle in which the origin lies ; in such a case, the angle bisector of the angle containing P is the same as that of the one containing the origin. P (x, y) L1 Origin L2 Fig - 25 To determine the angle bisector of the angle containing (1, –3), we simply determine the angle bisector of the angle not containing the origin, i.e. − x − 2 y + 11 −3 x + 6 y + 5 =− 5 3 5 ⇒ 6 5 x = 38 5 19 ⇒ x= 3 Note that to determine the angle bisector of the angle containing the point P as in Fig.-25, we would have chosen the angle bisector of the angle containing the origin. Maths / Straight Lines LOCUS 29 Art 10 Polar / Distance form of a line Sometimes, it is very convenient to write the equation of a straight line in polar / distance form. Suppose we know that the line passes through the fixed point P (h, k ) and is at an inclination of θ : y Q(x, y) r Let PQ = r P(h, k) θ x Fig - 26 For any point Q ( x , y ) at a distance r from P along this line, we can write the simple relation x−h y−k = =r cos θ sin θ This is the required equation of the line. The point Q ( x , y ), at a distance r from P, has the coordinates Q ( x, y ) ≡ ( h + r cos θ, k + r sin θ). Obviously, there will be another point, say Q′( x, y ), at a distance r from P along this line but on the opposite side of Q; thus Q′( x, y ) will have the coordinates Q′( x, y ) ≡ (h − r cos θ, k − r sin θ) Example – 15 A line through A( −5, − 4) meets the lines x + 3 y = 2, 2 x + y + 4 = 0 and x − y − 5 = 0 at the points B, C and D respectively. If 2 2 2 15 10 6 + = , AB AC AD find the equation of the line. Maths / Straight Lines LOCUS 30 Solution: B x-y-5=0 We want to C x + 3y = 2 find the equation of the line L. Assume AB = r1 A AC = r2 (-5,-4) 2x + y + 4 = 0 AD = r3 D L=0 Fig - 27 The figure above roughly sketches the situation described in the equation. Let B, C and D be at distances r1 , r2 and r3 from A along the line L = 0, whose equation we wish to determine. Assume the inclination of L to be θ. Thus, B, C and D have the coordinates (respectively): B ≡ (−5 + r1 cos θ, − 4 + r1 sin θ) C ≡ (−5 + r2 cos θ, − 4 + r2 sin θ) D ≡ (−5 + r3 cos θ, − 4 + r3 sin θ) Since these three points(respectively) satisfy the three given equations, we have : 15 Point B : (−5 + r1 cos θ) + 3(−4 + r1 sin θ) + 2 = 0 ⇒ r1 = cos θ + 3sin θ 10 Point C : 2( −5 + r2 cos θ) + ( −4 + r2 sin θ) + 4 = 0 ⇒ r2 = 2 cos θ + sin θ 6 Point D : (−5 + r3 cos θ) − (−4 + r3 sin θ) − 5 = 0 ⇒ r3 = cos θ − sin θ It is given that 2 2 2 15 10 6 + = AB AC AD 2 2 2 15 10 6 i.e. + = r1 r2 r3 ⇒ (cos θ + 3sin θ)2 + (2 cos θ + sin θ) 2 = (cos θ − sin θ) 2 ⇒ 4 cos 2 θ + 9 sin 2 θ + 12 sin θ cos θ = 0 ⇒ (2 cos θ + 3sin θ)2 = 0 −2 ⇒ tan θ = 3 −2 ⇒ m= 3 Maths / Straight Lines LOCUS 31 −2 Thus, we obtain the slope of L as . The equation of L can now be easily written : 3 −2 L : y − ( −4) = ( x − (−5)) 3 ⇒ L : 2 x + 3 y + 22 = 0 Maths / Straight Lines LOCUS 32 TRY YOURSELF - I x y x y 1. A variable straight line drawn through the intersection of the lines + = 1 and + = 1 meets the a b b a axes in A and B. Show that the locus of the mid-point of AB is 2 xy ( a + b) = ab( x + y ) 2. The line bx + ay = ab cuts the axes in A and B. Another variable line cuts the axes in C and D such that OA + OB = OC + OD, where O is the origin. Prove that the locus of the point of intersection of the lines AD and BC is the line x + y = a + b. 3. A point P moves so that the square of its distance from (3, –2) is equal to its distance from the line 5 x − 12 y = 13. Find the locus of P. 4. A line intersects the x-axis in A(7, 0) and the y-axis in B(0, –5). A variable line perpendicular to AB intersects the x-axis in P and the y-axis in Q. If AQ and BP intersect in R, find the locus of R. 5. If the sum of the distances of a point from two perpendicular lines in a plane is 1, prove that its locus is a square. 6. A vertex of an equilateral triangle is (2, 3) and the opposite side is x + y = 2. Find the equations of the other sides. 7. A ray of light along the line x − 2 y − 3 = 0 is incident upon the mirror-line 3 x − 2 y − 5 = 0. Find the equation of the reflected ray. 8. If the vertices of a triangle have integral coordinates, show that it cannot be equilateral. 9. Show using coordinate geometry that the angle bisectors of the sides of a triangle are concurrent. 10. The sides of a triangle are 4 x + 3 y + 7 = 0, 5 x + 12 y − 27 = 0 and 3 x + 4 y + 8 = 0. By explicitly evaluating the medians in this triangle, show that they are concurrent. 11. A rod APB of constant length meets the axes in A and B. If AP = b and PB = a and the rod slides between the axes, show that the locus of P is b 2 x 2 + a 2 y 2 = a 2b 2 12. If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and 1 1 1 b, show that 2 = 2 + 2. p a b 13. The lines 3 x + 4 y − 8 = 0 and 5 x + 12 y + 3 = 0 intersect in A. Find the equations of the lines passing through P (3, 4), which intersect the given lines at B and C, such that AB = AC. 14. The equal sides AB and AC of an isosceles triangle ABC are produced to the points P and Q such that BP ⋅ CQ = AB 2 . Prove that the line PQ always passes through a fixed point. 15. One side of a square is inclined to the x-axis at an angle α and one of its extremities is at the origin; prove that the equations to its diagonals are y (cos α − sin α ) = x(sin α + cos α ) and y (sin α + cos α ) + x (cos α − sin α ) = a where a is the length of the side of the sqaure. Maths / Straight Lines LOCUS 33 Section - 3 PAIR OF STRAIGHT LINES Consider two lines L1 ≡ y − m1 x − c1 = 0 L2 ≡ y − m2 x − c2 = 0 What do you think will L1L2 = 0 represent ? It is obvious that any point lying on L1 and L2 will satisfy L1L2 = 0, and thus L1L2 = 0 represents the set of points constituting both the lines, i.e., L1 L2 = 0 represents the pair of straight lines given by L1 = 0 and L2 = 0 For example, consider the equation y 2 − x 2 = 0. What does this represent ? We have y2 − x2 = 0 ...(1) ⇒ ( y + x)( y − x) = 0 ⇒ (1) represents the pair of straight lines x = y and x + y = 0. y x=y x The pair of lines given 2 2 by y - x = 0 x+y=0 Fig - 28 There is nothing special about considering a pair. We can similarly define the joint equation of n straight lines Li ≡ y − mi x − ci = 0 (i = 1, 2..., n) as L1L2 ...Ln = 0 ⇒ ( y − m1 x − c1 )( y − m2 x − c2 )...( y − mn x − cn ) = 0 ...(2) Any point lying on any of these n straight lines will satisfy (2), and thus (2) represents the set of all points constituting the n lines, i.e. (2) represents the joint equation of the n straight lines. What is relevant to us at this stage is only a pair of straight lines and it is on a pair of lines that we now focus our attention. Maths / Straight Lines LOCUS 34 PAIR OF LINES PASSING THROUGH THE ORIGIN We first consider a special (and simple)case. Both the lines in our pair pass through the origin. Thus, their equations can be written as L1 : y − m1 x = 0 L2 : y − m2 x = 0 y L2 θ L1 θ2 Pair of lines passing through the origin with slopes O θ1 x m1 = tan θ1 m2 = tan θ2 Fig - 29 The joint equation of this pair is : L1L2 = 0 ⇒ ( y − m1 x)( y − m2 x) = 0 ⇒ y 2 + m1m2 x 2 − (m1 + m2 ) xy = 0 ...(3) (3) suggests that the general equation of a pair of straight lines passing through the origin is ax 2 + 2hxy + by 2 = 0 ...(4) (4) is a homogenous equation of degree 2, implying that the degree of each term is 2. It should now be apparent that any homogenous equation of degree 2 will represent two straight lines passing through the origin (we’ll soon see that the two straight lines might be imaginary; the meaning of this will become clear in a subsequent example). Generalising, any nth degree homogenous equation of the form a0 x n + a1 x n −1 y + a2 x n − 2 y 2 + ... + an y n = 0 ...(5) represents n straight lines (real or imaginary) passing through the origin. To obtain the slopes of these n lines, we y divide by x n in (5) and substitute = m : x an m n + an −1m n −1 + ... + a0 = 0 The n values of m gives us the slopes of the n lines. Maths / Straight Lines LOCUS 35 Example – 16 Find the straight lines represented by (a) y 2 − 5 xy + 6 x 2 = 0 (b) 3 y 2 − 10 xy + 3 x 2 = 0 (c) y 2 + xy + x 2 = 0 Solution: Note that the homogenous nature of these equations tells us that the lines will pass through the origin. (a) y 2 − 5 xy + 6 x 2 = 0 We either factorize this equation straightaway : ( y − 2 x )( y − 3 x ) = 0 so that the lines are y = 2 x and y = 3 x OR, y we divide it by x 2 and substitute m = to obtain : x m 2 − 5m + 6 = 0 ⇒ m = 2, 3 y ⇒ = 2, 3 x ⇒ y = 2 x or y = 3 x Both alternatives are entirely equivalent. (b) 3 y 2 − 10 xy + 3 x 2 = 0 y ⇒ 3m − 10m + 3 = 0 2 m = x 1 ⇒ m = 3, 3 x ⇒ y = 3 x, y = 3 Maths / Straight Lines LOCUS 36 (c) y 2 + xy + x 2 = 0 y ⇒ m2 + m + 1 = 0 Again, m = x This has no real roots and thus physically, no lines will exist with the joint equation y 2 + xy + x 2 = 0. We sometimes say that this equation represents imaginary lines. Note that in the entire plane, only (0, 0) satisfies this equation. ____________________________________________________________________________________ Consider now that we’ve been given the equation of a pair of straight lines passing through the origin as : ax 2 + 2hxy + by 2 = 0 ...(1) We wish to determine the angle between these two lines. Let m1 and m2 be the slopes of these two lines. By y dividing(1) by x2 and substituting = m, we have x bm 2 + 2 hm + a = 0 This quadratic in m will have its roots as m1 and m2. Thus, −2h a m1 + m2 = ; m1m2 = ...(2) b b The angle between the two lines, say θ, is given by m1 − m2 tan θ = 1 + m1m2 (m1 + m2 ) 2 − 4m1m2 = 1 + m1m2 2 h 2 − ab = (Using (2)) a+b As a consequence of this formula, we see that the lines represented by (1), are : Parallel (in fact coincident since both pass through the origin) if h 2 = ab Perpendicular if a+b = 0 The importance of this condition must be mentioned; it is very widely used and should be committed to memory. As an example, the locus given by 3 y 2 − 8 xy − 3 x 2 = 0 ...(3) represent two perpendicular straight lines since a + b = (3) + ( −3) = 0 Verify this by the explicit factorization of (3). Maths / Straight Lines LOCUS 37 Example – 17 Find the equation of the pair of lines through the origin and perpendicular to the pair of lines ax 2 + 2hxy + by 2 = 0. Solution: Let the slopes of the two lines represented by the given equation be m1 and m2. As explained earlier, m1 and m2 are the roots of the quadratic bm 2 + 2 hm + a = 0 so that −2h a m1 + m2 = , m1m2 = ...(1) b b −1 −1 The slopes of the lines whose joint equation we require will simply be m and m so that their 1 2 equations will be : −1 −1 y= x, y= x m1 m2 ⇒ x + m1 y = 0, x + m2 y = 0 The required joint equation is ( x + m1 y )( x + m2 y ) = 0 ⇒ x 2 + m1m2 y 2 + (m1 + m2 ) xy = 0 a 2 2h ⇒ x2 + y − xy = 0 (Using (1)) b b ⇒ bx 2 − 2hxy + ay 2 = 0 Example – 18 he equation ax 3 + bx 2 y + cx 2 + dy 3 = 0 is a third degree homogenous equation and hence represents three straight lines passing through the origin. Find the condition so that two of these three lines may be perpendicular. y Solution: We divide the given equation by x 3 and substitute = m to obtain: x dm3 + cm 2 + bm + a = 0 ...(1) This has three roots, say m1 , m2 , m3 , corresponding to the three straight lines. Since we want two of these lines to be perpendicular, we can assume m1m2 = −1 Maths / Straight Lines LOCUS 38 From (1), we have −a m1 m2 m3 = d a ⇒ m3 = d Substituting this value of m3 back in (1), (since m3 is a root of (1)), we obtain da 3 ca 2 ba + 2 + +a =0 d3 d d ⇒ a 2 + ac + bd + d 2 = 0 Example – 19 Find the area of the triangle formed by the lines y 2 − 9 xy + 18 x 2 = 0 and y = 9. Solution: The joint equation can be factorized to obtained ( y − 3 x )( y − 6 x ) = 0 Thus, the three lines forming the sides of the triangle are y = 3 x, y = 6 x, y = 9 The three intersection points can easily be seen to be 3 (0, 0), (3, 9), ,9 2 Thus, the area of the triangle is 3 0 3 2 1 ∆= 0 9 9 2 1 1 1 27 = sq. units 4 Example – 20 The slope of one of the two lines represented by ax 2 + 2hxy + by 2 = 0 is the square of the other. Prove that a + b 8h 2 + =6 h ab Maths / Straight Lines LOCUS 39 Solution: Let the two slopes be m and m2; these are the roots of the quadratic bM 2 + 2 hM = a = 0 so that −2 h a m + m2 = , m3 = b b Cubing the first relation, we have −8h3 m3 + m6 + 3m3 (m + m 2 ) = b3 a a 2 3a −2h −8h3 ⇒ + + = 3 b b2 b b b ⇒ ba 2 + ab 2 + 8h 3 = 6hab a + b 8h 2 ⇒ + =6 h ab _____________________________________________________________________________________ GENERAL EQUATION OF A PAIR OF LINES Consider the equations of two arbitrary lines L1 and L2: L1 : l1 x + m1 y + n1 = 0 L2 : l2 x + m2 y + n2 = 0 The joint equation of the two lines is L1L2 = 0 ⇒ (l1 x + m1 y + n1 )(l2 x + m2 y + n2 ) = 0 ⇒ (l1l2 ) x 2 + (l1m2 + m1l2 ) xy + (m1m2 ) y 2 + (n1l2 + n2l1 ) x + (n1m2 + n2 m1 ) y + n1n2 = 0 ...(1) (1) suggests that the most general equation to a pair of straight lines has the form ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 ...(2) It might be apparent to you that (2) will not always represent a pair of straight lines. For (2) to indeed represent a pair of straight lines, it must be able to be factorised into two linear factors; as an exercise for the reader, show that (2) can be expressed as a product of linear factors if the following condition is satisfied: abc + 2 fgh − af 2 − bg 2 − ch 2 = 0 a h g ⇒ h b f =0 g f c Maths / Straight Lines LOCUS 40 We now re-evaluate the conditions for parallel and perpendicular lines, in the general case : Let L1 and L2 be two lines with slopes s1 and s2; their equations have already been mentioned above. L1 and L2 are parallel if s1 = s2 −l1 −l2 ⇒ = m1 m2 ⇒ l1m2 = l2 m1 ⇒ (l1m2 − l2 m1 ) =0 2 ⇒ (l1m2 + l2 m1 ) = 4 l1 l2 m1 m2 2 ...(3) From (2), observe that the constraint in (3) can be specified as 4h 2 = 4ab ⇒ h 2 = ab : Parallel lines This is the same condition as the one for the homogenous case. For L1 and L2 to be perpendicular, s1 ⋅ s2 = −1 −l1 −l2 ⇒ ⋅ = −1 m1 m2 ⇒ l1l2 + m1m2 = 0 ⇒ a+b = 0 : Perpendicular lines Again, this condition is the same as the one in the homogenous case. If fact, you can verify that the angle θ subtended between the two lines is also given by the same formula as in the homogenous case, i.e., 2 h 2 − ab tan θ = a+b That these formulae in the homogenous and the general case are the same should be obvious since the slope of any line is independent of the constant term appearing in its equation. Example – 21 Prove that the equation 6 x 2 + 13 xy + 6 y 2 + 8 x + 7 y + 2 = 0 represents a pair of straight lines. Find the point of intersection and the angle between these two lines. Maths / Straight Lines LOCUS 41 Solution: To show that this equation represents a pair of straight lines, we use the determinant condition mentioned earlier: 13 6 4 a h g 2 13 7 h b f = 6 2 2 g f c 7 4 2 2 49 13 91 = 6 12 − + (14 − 13) + 4 − 24 4 2 4 3 13 = − + −5 2 2 =0 which confirms the stated assertion. The angle between these two lines is given by 2 h 2 − ab tan θ = a+b 5 = 12 5 ⇒ θ = tan −1 12 To find the point of intersection, we must factorise the joint equation to obtain the separate equations of the lines. This task can be made easy be observing that since the homogenous part of the given equation is 6 x 2 + 13 xy + 6 y 2 = 0 which can be factorised as (2 x + 3 y )(3 x + 2 y ) = 0, the actual factors of the (original) equation will be of the form (2 x + 3 y + α ) (3 x + 2 y + β) = 0 Convince yourself about this argument. α and β can easily be evaluated using comparison of coefficients to be 2 and 1 respectively. Thus, the two lines are L1 : 2 x + 3 y + 2 = 0 L2 : 3x + 2 y + 1 = 0 1 −4 so that their point of intersection is, by solving this system of equations, , . 5 5 Maths / Straight Lines LOCUS 42 Example – 22 Show that the four lines given by the equations 3 x 2 + 8 xy − 3 y 2 = 0 3 x 2 + 8 xy − 3 y 2 + 2 x − 4 y − 1 = 0 form a square. What is the length of the sides of the square ? Solution: The first joint equation can be easily factorised to yield (3 x − y )( x + 3 y ) = 0 ⇒ 3 x − y = 0, x + 3y = 0 ...(1) These are perpendicular lines intersecting at the origin. As described in the previous example, the second joint equation can be factorised as (3 x − y + α ) ( x + 3 y + β) = 0 where α and β can be determined by the comparison of coefficients : α = −1, β = 1 The other two sides are thus 3 x − y − 1 = 0, x + 3 y + 1 = 0 ...(2) From (1) and (2), it should be evident that the four lines form a square. The length l of the sides of this square can be evaluated by determining the perpendicular distance between any pair of opposite sides, say 3 x − y = 0 and 3 x − y − 1 = 0 : 0 − (−1) 1 l= = 12 + 32 10 Example – 23 Find the joint equation of the angle bisectors of the lines represented by ax 2 + 2hxy + by 2 = 0. Solution: Let the slopes of the two lines represented by ax 2 + 2hxy + by 2 = 0 be m1 and m2, so that m1 and m2 are the roots of the quadratic bm 2 + 2 hm + a = 0 Thus, −2h a m1 + m2 = , m1m2 = b b Maths / Straight Lines LOCUS 43 It should be obvious that the angle bisectors will also pass through the origin, as shown below: y L2 A2 A1 L1 L1 and L2 are the original φ2 θ2 pair of lines while A1 and A2 θ1 φ1 are their angle bisectors x m1 = tanθ1 m 2 = tanθ2 Fig - 30 From the figure, observe that θ1 + θ2 π θ1 + θ2 φ1 = , φ2 = + 2 2 2 ⇒ 2φ1 = θ1 + θ2 , 2φ2 = π + θ1 + θ2 ⇒ tan 2φ1 = tan 2φ2 = tan(θ1 + θ2 ) tan θ1 + tan θ 2 = 1 − tan θ1 tan θ2 m1 + m2 = 1 − m1m2 2h = ... (1) a −b Now, for any point ( x, y ) on the angle bisector A1 (or A2 ), y y tan φ1 = or tan φ2 = x x 2 xy We have used φ to ⇒ tan 2φ = represent both φ and φ x − y2 2 1 2 From this relation and (1), we have x 2 − y 2 xy = a−b h Maths / Straight Lines LOCUS 44 This is the joint equation of the angle bisectors; as expected, it is a second degree homogenous equation. As a corollary, suppose we are required to find the joint equation of the angle bisectors of the lines L1 and L2 represented by ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 We first find the point of intersection of L1 and L2, say P (α, β). If we now shift our coordinate system (translation) so that P is the origin (Refer to Appendix - 1), and denote the coordinates in the new system by ( X , Y ), we will have the joint equation of the angle bisectors of L1 and L2 as X 2 − Y 2 XY = a −b h But since X → x − α, Y → y − β, the joint equation in the original frame is ( x − α) 2 − ( y − β) 2 ( x − α)( y − β) = a −b h ____________________________________________________________________________________ We now discuss a very useful application of the concept of pair of straight lines. Consider a second degree curve S ( x, y ) with the equation S ( x, y ) ≡ ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 and a straight line L ≡ px + qy + r = 0 intersecting S = 0 in A and B. Let O be the origin. What is the joint equation of OA and OB ? y B What is the joint equation of the pair of straight lines A OA and OB? x O L=0 S(x, y) =0 Fig - 31 Maths / Straight Lines LOCUS 45 The insight that we use here is that since both OA and OB pass through the origin, their joint equation will be homogenous. We now construct a homogenous equation and show that both A and B satisfy it; that equation is then guaranteed to jointly represent OA and OB. First of all, observe that since A and B satisfy the equation of L, i.e. px + qy + r = 0, they will also satisfy the relation px + qy =1 : Both A and B will satisfy this relation. −r Now, we homogenize the equation of the second degree curve S ( x, y ) using the relation above; consider this equation : px + qy px + qy px + qy 2 ax + 2hxy + by + 2 gx 2 2 + 2 fy + c =0 ...(1) −r −r −r Can you understand why we’ve done this? The equation we obtain above is a second degree homogenous equation, and so it must represent two straight lines passing through the origin. Which two straight lines? Since A and B px + qy satisfy the equation of the original curve as well as the relation = 1, A and B both satisfy the homogenized −r equation in (1). What does this imply ? That (1) is the joint equation of OA and OB! Go over this discussion again if you find this confusing. You must fully understand the described technique which will find very wide usage in subsequent chapters. Example – 24 Find the joint equation of the straight lines passing through the origin O and the points of intersection of the line 3 x + 4 y − 5 = 0 and the curve 2 x 2 + 3 y 2 = 5. Solution: One approach is of course to explicitly determine the two points of intersection, say A and B, writing the equations of OA and OB, thereby obtaining the required joint equation. You are urged to do this as an exercise. However, we’ll use the homogenizing technique just described : y B A We wish to determine x the joint equation of O OA and OB 2 2 2x + 3y = 5 Fig - 32 Maths / Straight Lines LOCUS 46 The required equation can be obtained by first writing the line as 3x + 4 y =1 5 and then using this relation to homogenize the equation of the curve : 3x + 4 y 2 2x + 3 y = 5 2 2 5 ⇒ 10 x 2 + 15 y 2 = 9 x 2 + 16 y 2 + 24 xy ⇒ x 2 − 24 xy − y 2 = 0 This is the required joint equation of OA and OB. Example – 25 Find the value of m, if the lines joining the origin O to the points of intersection A, B of y = 1 + mx and x 2 + y 2 = 1 are perpendicular. Solution: The joint equation of OA and OB is x 2 + y 2 = ( y − mx) 2 ⇒ (1 − m2 ) x 2 + 2mxy = 0 ... (1) The condition for perpendicularity is a+b = 0 which when applied to (1) yields 1 − m2 = 0 m = ±1 This example was more or less trivial and a little knowledge of circles would have enabled you to solve this question without resorting to the homogenizing approach; however, the fact that this approach is very powerful in many cases will become apparent in later examples. Maths / Straight Lines LOCUS 47 TRY YOURSELF - II Q. 1 Find the values(s) of m for which the following equation(s) represents a pair of straight lines: (a) x 2 + λxy − 2 y 2 + 3 y − 1 = 0 (b) 4 x 2 + 10 xy + λy 2 + 5 x + 10 y = 0 Q. 2 Find the angle of intersection of the straight lines given by the equation 3 x 2 − 7 xy + 2 y 2 + 9 x + 2 y − 12 = 0 Q. 3 Show that the lines joining the origin to the points common to x 2 + hxy − y 2 + gx + fy = 0 and fx − gy = λ are at right angles for all values of λ. Q. 4 Find the angle between the lines given by x 2 − 2 pxy + y 2 = 0. x y Q. 5 Prove that the lines joining the origin to the points of intersection of the line + = 2 with the curve h k ( x − h) 2 + ( y − k ) 2 = c 2 , are perpendicular if h 2 + k 2 = c 2 . Q. 6 Find the joint equations of the straight lines passing through (1, 1) and parallel to the lines given by x 2 − 5 xy + 4 y 2 + x + 2 y − 2 = 0. Q. 7 Evaluate the point of intersection for the lines represented by the general equation ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0. Q. 8 Find the joint equation of the images of the pair of lines ax 2 + 2hxy + by 2 = 0 in the mirror y = 0. Q. 9 Find the joint equation of the angle bisectors of the lines given by x 2 + 2 xy sec θ + y 2 = 0. Q. 10 If the pairs of straight lines ax 2 + 2hxy + by 2 = 0 a′x 2 + 2h′xy + b′y 2 = 0 have a line in common, show that ab′ − a′b 2 = ( ha′ − h′a )( h′b − hb′) 2 Maths / Straight Lines LOCUS 48 SOLVED EXAMPLES Example – 1 Consider a fixed point O and n fixed straight lines. Through O, a (variable) line is drawn intersecting the fixed lines in P , P2 ....., Pn . On this variable line, a point P is taken such that 1 n 1 1 1 = + + ..... + . OP OP OP2 1 OPn Find the locus of P. Solution: P1 P2 P3 Pn A figure illustrating O ... the situation described L1 L2 L3 L n Fig - 33 Assume the equations of the fixed lines to be Li ≡ ai x + bi y + ci = 0, i = 1, 2,.....n and the coordinates of the fixed point O to be ( h, k ). Let the slope of the variable line be represented by tan θ. Thus, the points Pi have the coordinates Pi ≡ (h + OPi cos θ, k + OPi sin θ) i = 1, 2.....n Since each Pi satisfies Li , we have ai (h + OPi cos θ) + bi (k + OPi sin θ) + ci = 0 −( hai + kbi + ci ) ⇒ OPi = ai cos θ + bi sin θ Assume the coordinates of P (whose locus we wish to determine) to be ( x, y ). Thus, we have x = h + OP cos θ, y = k + OP sin θ ...(1) Maths / Straight Lines LOCUS 49 From the relation given in the equation, we have n 1 =∑ NP OPi n a cos θ + bi sin θ ⇒ = −∑ i NP hai + kbi + ci ai bi = −∑ cos θ + −∑ sin θ hai + kbi + ci hai + kbi + ci These substitutions have = λ cos θ + µ sin θ been doen for convenience From (1), we have n λ ( x − h) µ ( y − k ) = + OP OP OP ⇒ λ x + µy − ( λ h + µk + n ) = 0 This is the locus of the point P; it is evidently a straight line. Example – 2 Lines are drawn to intersect n concurrent lines at the points A1 , A2 ....., An such that n 1 ∑ OA i =1 = constant i where O is the point of concurrency. Show that the variable lines all pass through a fixed point. Solution: There’s no loss of generality in assuming O to be the origin since we are dealing only with lengths which are invariant with respect to the choice of the coordinate axes. Ln L2 L1 L A2 Li's are all fixed lines. A1 L is the variable line O Fig - 34 Maths / Straight Lines LOCUS 50 The inclinations of the fixed lines can be assumed to be θi so that the points Ai have the coordinates Ai ≡ (OAi cos θi , OAi sin θi ) Let the variable line have the equation ax + by + c = 0 Since all the Ai ' s lie on this line, we have aOAi cos θi + bOAi sin θi + c = 0 −c ⇒ OAi = ...(1) a cos θi + b sin θi According to the condition specified in the question, n 1 ∑ OA i =1 = costant = K (say) ...(2) i Thus, using (1) in (2), we have n a cos θi + b sin θi ∑ i =1 −c =K n n ∑ cos θi ∑ sin θi a i =1 + b i =1 +c = 0 K K ...(3) n n ∑ cos θi ∑ sin θ i (3) shows that the variable line L always passes through the fixed point i =1 , i =1 . K K Example – 3 Prove that the centroid G of a triangle divides the line joining its circumcentre C and its orthocentre H in the ratio 1 : 2. Maths / Straight Lines LOCUS 51 Solution: To make our task simpler, we choose a coordinate frame in which the triangle’s vertices have coordinates that are “easy” to work with. One such choice is shown below. y R(b,c) x P(-a,0) Q(a,0) Fig - 35 Now, we find G, C and H : −a + a + b b The centroid G(xG , yG ): xG = = 3 3 0+0+c c yG = = 3 3 b c ⇒ G≡ , 3 3 The circumcentre C(xC , yC ) : To find C, we need two perpendicular bisectors ( ⊥ B ). The ⊥ B of PQ is x=0 c Since the slope of PR is , the equation of the ⊥ B b+a of PR is c b + a b−a y− = − x − 2 c 2 The two ⊥ Bs intersect at C: b2 − a2 + c2 C ≡ 0, 2c Maths / Straight Lines LOCUS 52 The orthocentre H(x H , y H ) : The altitude from R onto PQ is simply x=b Let find the altitude from Q onto PR: b+a y −0 = − ( x − a) c The two altitudes intersect at b2 − a 2 H ≡ b, − c The point which divides CH in the ratio 1 : 2 is b2 − a 2 b2 − a2 + c2 1 × − + 2× 1× b + 2 × 0 , c 2c ≡b, c 3 3 3 3 which is the same as G. Thus, G divides CH in the ratio 1 : 2. Example – 4 Find the area of the parallelogram formed by the lines a1 x + b1 y + c1 = 0; a1 x + b1 y + d1 = 0 a2 x + b2 y + c2 = 0; a2 x + b2 y + d2 = 0 Find also the condition for this parallelogram to be a rhombus. Solution: a1x + b1y + d1 = 0 d2 = 0 c2 = 0 We need to find the b2 y + b2 y + area of this parallelogram. a2 x + a2 x + α a1x + b1y + c1 = 0 Fig - 36 Maths / Straight Lines LOCUS 53 We first consider a little geometry for this parallelogram. Let the parallelogram have sides a and b and let the perpendicular distances between its opposite sides be p1 and p2: p1 b p2 α a Fig - 37 Then, the area A of the parallelogram is A = ab sin α = ap1 (∵ p1 = b sin α) p2 = ⋅ p1 (∵ p2 = a sin α) sin α p1 p2 = ...(1) sin α Thus, the area of the parallelogram can be expressed using the perpendicular distances between its opposite sides rather than using the length of the sides. This is good for us since we already know how to evaluate the perpendicular distance between two parallel lines. c1 − d1 ⇒ p1 = ...(2) a12 + b12 c2 − d 2 ⇒ p2 = ...(3) a2 + b22 2 Also, a1 a2 − −− tan α = b1 b2 aa 1+ 1 2 b1b2 a1b2 − a2b1 = a1a2 + b1b2 Maths / Straight Lines LOCUS 54 so that a1b2 − a2 b1 sin α = ...(4) (a12 + b12 )(a2 + b22 ) 2 Using (2), (3) and (4) in (1), we have (c1 − d1 )(c2 − d 2 ) A= (a1b2 − a2 b1 ) Now, the parallelogram is a rhombus if its diagonals are perpendicular, which also means that the distances between its opposite sides are equal, i.e. p1 = p2 c1 − d1 c2 − d 2 ⇒ = a12 + b12 a2 + b22 2 Example – 5 A rod AB of length l slides with its end on the coordinate axes. Let O be the origin. The rectangle OAPB is completed. Find the locus of the foot of the perpendicular drawn from P onto AB. Solution: y A P We need to find the locus of F as AB slides between F the axes l θ x O B Fig - 38 In terms of the variable θ, A and B, and hence P, have coordinates A ≡ (0, l sin θ), B ≡ (l cos θ, 0), P ≡ (l cos θ, l sin θ) The slope of PF can be observed to be cot θ so that its equation is y − l sin θ = cot θ( x − l cos θ) ...(1) Maths / Straight Lines LOCUS 55 The equation of AB is x sec θ + y cosec θ = l ...(2) The intersection of (1) and (2) gives us the point F ( h, k ) : h = l cos3 θ, k = l sin 3 θ 1/ 3 1/ 3 h k ⇒ cos θ = , sin θ = l l Eliminating θ, we have h2 / 3 + k 2 / 3 = l 2 / 3 Thus, the locus of F is x2 / 3 + y 2 / 3 = l 2 / 3 Example – 6 (a) Consider a line segment AB where A ≡ ( x1 , y1 ) and B ≡ ( x2 , y2 ). In what ratio does a line L ≡ ax + by + c = 0 divide AB? Solution: Let the required ratio be λ :1 1 B(x2, y2) λ C A ( x 1 , y1 ) L = ax + by + c = 0 Fig - 40 The coordinates of C are (from the internal division formula), λx + x λy + y1 C ≡ 2 1, 2 λ +1 λ +1 Since this lies on L, we have λx + x λy + y a 2 1 + b 2 1 + c = 0 λ +1 λ +1 ⇒ λ(ax2 + by2 + c) + (ax1 + by1 + c) = 0 − ax1 + by1 + c ⇒ λ= ax2 + by2 + c Maths / Straight Lines LOCUS 56 L( x1 , y1 ) ⇒ λ=− L( x2 , y2 ) This is a useful result (as we’ll see from part(b), the next example) and it would be worth memorizing it. Example – 6 (b) A line intersects BC, CA and AB in ∆ABC at P, Q and R respectively. Show that BP CQ AR ⋅ ⋅ = −1 PC QA RB Solution: A ( x 1 , y1 ) L=0 Q B C (x 3 , y3 ) (x2, y2) P R Fig - 39 Using the result of the last example, we have AR L( x1 , y1 ) =− ...(1) RB L ( x2 , y 2 ) BP L( x2 , y2 ) =− ...(2) PC L( x3 , y3 ) CQ L( x3 , y3 ) =− ...(3) QA L ( x1 , y1 ) From (1), (2) and (3), it should be evident that the assertion stated in the question is valid. Example – 7 The curves C1 : a1 x 2 + 2h1 xy + b1 y 2 + 2 g1 x = 0 C2 : a2 x 2 + 2h2 xy + b2 y 2 + 2 g 2 x = 0 intersect at two points A and B other than the origin. Find the condition for OA and OB to be perpendicular. Maths / Straight Lines LOCUS 57 Solution: Assume the equation of AB to be y = mx + c. Thus, using the homogenizing technique, we can write the joint equation of OA and OB: y − mx Homogenizing C1: a1 x 2 + 2h1 xy + b1 y 2 + 2 g1 x =0 c 2mg1 2 2 g1 xy ⇒ a1 − x + 2h1 xy + b1 y + 2 =0 c c This is the joint equation of OA and OB. OA and OB are perpendicular if 2mg1 a1 − + b1 = 0 c m a1 + b1 ⇒ = ...(1) c 2 g1 Homogenizing C2: Similarly, we can again evaluate the joint equation of OA and OB by homogenizing the equation of C2 : 2mg 2 2 2 g 2 xy a2 − x + 2h2 xy + b2 y + =0 2 c c The perpendicularity condition gives 2mg 2 a2 − + b2 = 0 c m a2 + b2 ⇒ = ...(2) c 2g2 From (1) and (2), the necessary required condition is a1 + b1 a2 + b2 = g1 g2 Example – 8 Find the orthocentre of the triangle formed by the lines ax 2 + 2hxy + by 2 = 0 and px + qy = 1. Maths / Straight Lines LOCUS 58 Solution: The two lines given by the joint equation pass through the origin. Assume their slopes to be m1 and m2 so that m1 and m2 are the roots of bm 2 + 2 hm + a = 0 2h a ⇒ m1 + m2 = − , m1m2 = ...(1) b b y y = m2 x N y = m1 x M px +qy = 1 O x Fig - 41 To evaluate the orthocentre, we need two altitudes. We take one of them to be the one dropped from O onto MN. qx − py = 0 ...(2) Let us now find the altitude from M onto ON. The coordinates of M are, by solving y = m1 x and px + qy = 1 simultaneously, 1 m1 M ≡ , p + qm1 p + qm1 1 The slope of ON is m2 so that the slope of the altitude through M is − m ; thus, its equation is 2 m1 −1 1 y− = x− p + qm1 m2 p + qm1 x 1 + m1m2 ⇒ +y= m2 m2 ( p + qm1 ) 1 + m1m2 ⇒ x + m2 y = ...(3) p + qm1 Maths / Straight Lines LOCUS 59 The intersection of (2) and (3) yields the orthocentre ( h, k ) : p (1 + m1m2 ) h= ( p + qm1 )( p + qm2 ) p (1 + m1m2 ) = p + pq ( m1 + m2 ) + q 2 m1m2 2 p (a + b) = bp − 2hpq + aq 2 2 q k= .h (from (1)) p q ( a + b) = bp − 2hpq + aq 2 2 Thus, the orthocentre has the coordinates p ( a + b) q ( a + b) 2 , 2 bp − 2hpq + aq bp − 2hpq + aq 2 2 Example – 9 Show that the equation λ( x3 − 3 xy 2 ) + y 3 − 3 x 2 y = 0 represents three straight lines equally inclined to one another. Solution: Observe that since the equation is homogenous, it will represent three straight lines passing through the origin. Let the slopes of the three lines be m1 , m2 and m3 . Thus m1 , m2 and m3 are the roots of the equation y λ(1 − 3m 2 ) + m3 − 3m = 0 where m = x 3m − m3 ⇒ =λ 1 − 3m2 Maths / Straight Lines LOCUS 60 y Since m = = tan θ, where θ is the inclination of the line, we have x 3 tan θ − tan 3 θ λ= = tan 3θ 1 − 3 tan 2 θ ⇒ tan 3θ = λ ⇒ 3θ = nπ + tan −1 λ nπ + tan −1 λ ⇒ θ= 3 Since there are three lines corresponding to the joint equation, we’ll have three corresponding angles of inclination tan −1 λ π 2π θ1 = , θ2 = + tan −1 λ, θ3 = + tan −1 λ 3 3 3 The angles of inclination show that the three lines are equally inclined to one another. y L2 L3 π 3 π θ3 3 θ2 L1 π θ1 3 x Fig - 42 v Example – 10 Show that all the chords of the curve 3 x 2 − y 2 − 2 x + 4 y = 0 which subtend a right angle at the origin pass through a fixed point. Find that point. Maths / Straight Lines LOCUS 61 Solution: Let y = mx + c be a chord of the curve which subtends a right angle at the origin. The joint equation of the lines joining the origin to points of intersection of y = mx + c and the curve is y − mx 3 x 2 − y 2 + (4 y − 2 x) =0 c This represents two perpendicular lines if Coeff. of x 2 + Coeff. of y 2 = 0 2m 4 3+ + −1 = 0 c c ⇒ c+m+2 = 0 ⇒ ( −2) = m(1) + c ...(1) (1) shows that y = mx + c always passes through the fixed point (–2, 1). Maths / Straight Lines LOCUS 62 ASSIGNMENT [ LEVEL - I ] 1. Through the origin O, a (variable) line is drawn to cut the lines y = m1 x + c1 and y = m2 x + c2 at Q and R. Let there be a point P on this variable line such that OP is the geometric mean of OQ and OR. Find the locus of P. 2. Find the condition so that the pair of straight lines joining the origin to the points of intersection of y = mx + c and the circle x 2 + y 2 = a 2 may be perpendicular. 3. A line drawn through the origin intersects the lines 2 x + y − 2 = 0 and x − 2 y + 2 = 0 in A and B. Let M be the mid-point of AB. Show that the locus of M is 2 x 2 − 3 xy − 2 y 2 + x + 3 y = 0. 4. Show that the reflection of the line ax + by + c = 0, a ≠ b in the line x + y + 1 = 0 is the line bx + ay + ( a + b − c ) = 0. 5. Find the angle between the straight lines given by ( x 2 + y 2 )sin 2 α = ( x cos θ − y sin θ)2 . 6. Prove that the lines 2 x 2 + 6 xy + y 2 = 0 are equally inclined to the lines 4 x 2 + 18 xy + y 2 = 0. 7. Let ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represent a pair of parallel straight lines. Prove that the g 2 − ac f 2 − bc distance between these lines is d = 2 =2 . a ( a + b) b( a + b) 8. If the equation 2hxy + 2 gx + 2 fy + c = 0 represents two straight lines, show that they form a rectangle fg of area with the coordinate axes. h2 [ LEVEL - II ] 9. A straight line is such that the algebraic sum of the perpendiculars drawn upon it from any number of fixed points is zero. Show that it always passes through a fixed point. 10. Let a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 be two lines intersecting at P. Show that a1a2 + b1b2 < 0 ⇒ The angle of intersection at P containing the origin is acute. a1a2 + b1b2 > 0 ⇒ The angle of intersection at P containing the origin is obtuse. Maths / Straight Lines LOCUS 63 a a a 11. Show that there exists a point equidistant from the four points am1 , , am2 , , am3 , m1 m2 m3 a and , am1m2 m3 . m1m2 m3 12. The vertices of triangle are ( xi , xi tan θi ), i = 1, 2,3. The circumcentre of this triangle is the origin and its orthocentre is ( a, b). Show that a cos θ1 + cos θ2 + cos θ3 = . b sin θ1 + sin θ2 + sin θ3 13. A rectangle PQRS has its side PQ parallel to the line y = mx and the vertices P, Q and S lie on the lines y = a, x = b and x = −b respectively. Find the locus of R. 14. For points P ≡ ( x1 , y1 ) and Q ≡ ( x2 , y2 ), the M-distance d ( P, Q ) is defined as d ( P, Q) = x1 − x2 + y1 − y2 Let O ≡ (0, 0) and A ≡ (3, 2). Prove that the set of points in the first quadrant which are equi-M-distant from O and A, consists of the union of a line segment of finite length and an infinite ray. Sketch this set. 15. The sides of a triangle are Li ≡ x cos αi + y sin α i = pi , i = 1, 2,3. Prove that the orthocentre of this triangle satisfies L1 cos(α 2 − α3 ) = L2 cos(α3 − α1 ) = L3 cos(α1 − α 2 ). 16. If ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 and ax 2 + 2hxy + by 2 − 2 gx − 2 fy + c = 0, each represent 2c a pair of straight lines, prove that the area of the parallelogram they enclose is . h 2 − ab 17. Find the area of the triangle formed by the lines given by ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 and the x-axis. 18. If ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represents two straight lines, prove that the product of the c perpendiculars drawn from the origin to these lines is . ( a − b) 2 + 4h 2 19. A line L through the origin meets x + y = 1 and x + y = 3 at P and Q respectively. Through P and Q, two straight lines L1 and L2 are drawn parallel to 2 x − y = 5 and 3 x + y = 5 respectively. The lines L1 and L2 intersect in R. As L varies, show that the locus of R is a straight line. 20. Let the sides of a parallelogram be U = a, U = b, V = a′ and V = b′, where U = lx + my + n and V ≡ l ′x + m′y + n′. Show that the equation of the diagonal through the intersection points of U = a, V = a′ and U = b, V = b′ is given by U V 1 a a′ 1 = 0 b b′ 1 Maths / Straight Lines LOCUS 64 APPENDIX TRANSFORMATION OF COORDIN ATES TRANSFORMATION COORDINA Suppose that a person A is flying a kite from the ground while another person B is observing this kite from the top of a building, as shown below: y Y B (H, K) X A x (0,0) Fig - 44 In A's frame of reference, the coordinates of B are ( H , K ). Now suppose that A and B both specify the position of the kite relative to themselves. It should be evident that the coordinates of the kite in the two reference frames will be different. Let the coordinates of the kite be ( x, y ) in A's reference frame and ( X , Y ) in B's reference frame. Then, we have X = x− H, Y = y−K ...(1) Thus, a translation of the axes implies a corresponding change in the coordinates in the manner specified by (1). In fact, if the kite traces a path f ( x, y ) = 0 in A's reference frame, it will trace the path f ( X + H , Y + K ) = 0 in B's frame of reference. Maths / Straight Lines LOCUS 65 Translation of axes implies a simple shift in the origin without a change in the relative orientation of the axes. We now consider the case when the axes is rotated but the origin is the same y Y A X θ x Fig - 45 Let a point A have the coordinates ( x, y ) in the original frame of reference and ( X , Y ) in the rotated frame of reference. Verify that the following relations hold true : x = X cos θ − Y sin θ y = X sin θ + Y cos θ We can now combine the case of translation and rotation of axes to determine the most general transformed coordinates. Let the origin of the axes be shifted to ( h, k ) and the axes be rotated by an anticlockwise angle θ. The original coordinates ( x, y ) and the coordinates in the new frame of reference ( X , Y ) will satisfy the relations - x = h + X cos θ − Y sin θ y = k + X sin θ + Y cos θ Maths / Straight Lines

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