Straight Line Geometry PDF

8. STRAIGHT LINE 1. INTRODUCTION Co-ordinate geometry is the branch of mathematics which includes the study of different curves and figures by ordered pairs of real numbers called Cartesian co-ordinates, representing lines & curves by algebraic equation. This mathematical...

8. STRAIGHT LINE 1. INTRODUCTION Co-ordinate geometry is the branch of mathematics which includes the study of different curves and figures by ordered pairs of real numbers called Cartesian co-ordinates, representing lines & curves by algebraic equation. This mathematical model is used in solving real world problems. 2. CO-ORDINATE SYSTEM Co-ordinate system is nothing but a reference system designed to locate position of any point or geometric element in a plane of space. 2.1 Cartesian Co-ordinates Let us consider two perpendicular straight lines XOX’ and YOY’ passing through the origin y O in the plane. Then, =90 x Axis of x: The horizontal line xox’ is called axis of x. x’ O Axis of y: The vertical line yoy’ is called axis of y. y’ Co-ordinate axis: x-axis and y-axis together are called axis of co-ordinates or axis of Figure 8.1 reference. Origin: The point ‘O’ is called the origin of co-ordinates or just the origin. Oblique axis: When xox’ and yoy’ are not at right angle, i.e. if the both axes are not perpendicular, to each other, then axis of co-ordinates are called oblique axis. 2.2 Co-ordinate of a Point The ordered pair of perpendicular distances of a point from X- and Y-axes are called co-ordinates of that point. If the perpendicular algebraic distance of a point p from y-axis is x and from x-axis is y, then co-ordinates of the point P is (x, y). Here, (a) x is called x-co-ordinate or abscissa. (b) y is called y-co-ordinate or ordinate. (c) x-co-ordinate of every point lying upon y-axis is zero. (d) y-co-ordinate of every point lying upon x-axis is zero. (e) Co-ordinates of origin are (0, 0). Note: A point whose abscissa and ordinate are both integers is known as lattice point. 8. 2 | Straight Line 2.3 Polar Co-ordinates y P (r ,) Let OX be any fixed line, known as initial line, and O be the origin. If the distance of any point P from the origin O is ‘r’ and ∠XOP = θ, then (r, θ) are known as polar co-ordinates r of point P. If (x, y) are the Cartesian co-ordinates of a point P, then x = rcos θ; y = rsin θ and  x y | r | = x2 + y 2 , θ = tan–1   θ ∈ ( −π , π ) O x Figure 8.2 Illustration 1: If the Cartesian co-ordinates of any point are ( 3, 1) , find the polar co-ordinates.  (JEE MAIN) y Sol: Polar co-ordinates of any point are (r, θ), where r = x2 + y 2 and θ = tan–1  . x x= 3;y=1 Let their polar co-ordinates be (r, θ) ⇒ x = r cos θ; y = r sinθ So r ⇒ x2 + y 2 r = 3+1 y  1  π θ ⇒ tan−1   = 2 θ ⇒ tan−1  = x  3 6  π ∴(r, θ) =  2 , .  6 y 3. DISTANCE FORMULA Q(x2,y2) B(r2,2) d The distance between two points P(x1, y1) and Q(x2, y2) is ,y 1) P(x 1 PQ = (x1 – x2 )2 + (y1 – y 2 )2 = (x2 − x1 )2 + (y 2 − y1 )2 (r1,1) A x2 - x1 Distance of a point P(x1, y1) from the origin O(0, 0) is O x2 OP = 2 X +y 2 1 1 Figure 8.3 Distance between two polar co-ordinates A(r1, q1) and B(r2, q2) is given by AB = r12 + r22 − 2r1r2 cos(θ1 − θ2 ) Proof: AB = (x2 − x1 )2 + (y 2 − y1 )2 x1 = r1 cos q1, x2 = r2 cos q2, y1 = r1sin q1, y2 = r2sinq2 AB = (r2 cos θ2 − r1 cos θ1 )2 + (r2 sin θ2 − r1 sin θ1 )2 AB = (r2 cos θ2 )2 − 2r1r2 cos θ1 cos θ2 + (r1 cos θ1 )2 + (r2 sin θ2 )2 − 2r1r2 sin θ1 sin θ2 + (r1 sin θ1 )2 AB = r12 + r22 − 2r1r2 cos(θ1 − θ2 ) MASTERJEE CONCEPTS Distance between two polar co-ordinates A(r1, q1) and B(r2, q2) is given by AB = r12 + r22 − 2r1r2 cos(θ1 − θ2 ) Vaibhav Krishnan (JEE 2009, AIR 22)  M a them a tics | 8.3  π  π Illustration 2: Find the distance between P  2, −  and Q  3, .  (JEE MAIN)  6  6 Sol: The distance between two points = r12 + r22 − 2r1r2 cos(θ1 − θ2 ) Therefore,.  π π π 1 PQ = r12 + r22 − 2r1r2 cos(θ1 − θ2 ) = 4 + 9 − 2.2.3cos  − −  = = 4 + 9 − 12cos   = 13 − 12. = 7  6 6 3 2 Illustration 3: The point whose abscissa is equal to its ordinate and which is equidistant from the points A(1, 0), B(0, 3) is  (JEE MAIN) Sol: Given, abscissa = ordinate. Therefore distance can be found by considering the co-ordinates of required point be P(k, k). Now given PA = PB ⇒ (k − 1)2 + k 2 = k 2 + (k − 3)2 2k2 – 2k + 1 = 2k2 – 6k + 9 ⇒ 4k = 8 ⇒ k = 2 4. SECTION FORMULA Let R divide the two points P(x1, y1) and Q(x2, y2) internally in the ratio m:n. Let (x, y) be the co-ordinates of R. y Draw PM, QN, RK perpendicular to the x-axis. Q R F Also, draw PE and RF perpendicular to RK and QN. P E PR m Here, =. RQ n x’ O M K N x Triangles PRE and RFQ are similar. PR PE PE m y’ ∴ = ⇒ = RQ RF RF n Figure 8.4 But PE = x – x1 and RF = x2 – x. x − x1 m mx2 + nx1 ∴ = ⇒ x= x2 − x n m+n ER m In the same way, = FQ n y − y1 m my 2 + ny1  mx2 + nx1 my 2 + ny1  i.e., = ⇒y= The co-ordinates of R are  ,  y2 − y n m+n  m+n m + n  PR ' m If R’ divides PQ externally, so that = , triangles PER’ and QR’F are similar. y QR ' n PR ' PE F R’ ∴ = R 'Q R 'F Q P E But PE = x – x1 and R’F = x – x2. x x − x1 m mx2 − nx1 x’ O M N K ∴ = i.e., x = x − x2 n m−n my 2 − ny1 y’ Similarly, y =. m−n Figure 8.5  mx − nx1 my 2 − ny1  The co-ordinates of R’ are  2 ,   m−n m − n  8. 4 | Straight Line PR ' m m Alternate Method: =− = By changing n into –n in the co-ordinates of R, we can obtain the co-ordinates R 'Q n –n of R’: mx2 − nx1 my 2 − ny1 , m−n m−n  x + x 2 y1 + y 2  Cor. The mid-point joining the two points (x1, y1) and (x2, y2) is  1 ,   2 2 .  x + λx2 y1 + λy 2  Cor. From the above cor., the co-ordinates of a point dividing PQ in the ratio λ:1 are  1 , . Considering  1+ λ 1+ λ  λ as a variable parameter, i.e. of all values positive or negative, the co-ordinates of any point on the line joining the points (x1, y1) and (x2, y2) can be expressed in the above forms. 5. SPECIAL POINTS OF A TRIANGLE 5.1 Centroid C(x3, y3) Let the vertices of the triangle ABC be (x1, y1), (x2, y2) and (x3, y3), respectively. F 1 D  x + x3 y 2 + y 3  2 G The mid-point D of BC is  2 ,  G, the centroid, divides AD internally 2 2  in the ratio 2:1.  (x1, y1)A E B(x2, y2) Let G be (x, y), Figure 8.6 2. ( (x2 + x3 ) / 2 ) + 1.x1 x1 + x2 + x3 then x = = and 2+1 3 2. ( (y 2 + y 3 ) / 2 ) + 1.y1 y1 + y 2 + y 3  x + x 2 + x3 y1 + y 2 + y 3  y= = ∴ G is  1 , . 2+1 3  3 3  5.2 Incentre Let A (x1, y1), B (x2, y2), C (x3, y3) be the vertices of the triangle. Let AD bisect angle BAC and cut BC at D. A BD AB c We know that = = DC AC b cx3 + bx2 cy 3 + by 2 I Hence the co-ordinates of D are , C c+b c+b D Let (x, y) be the incentre of the triangle B CD b BC b + c ca AI AB c b+c = ∴ = ∴ BD = = = = Figure 8.7 BD c DB c b + c ID BD ( ca / (b + c) ) a (b + c) ( (cx3 + bx2 ) / (c + b) ) + ax1 ax1 + bx2 + cx3 = x = , b+c+a a+b+c (b + c) ( (cx3 + bx2 ) / (c + b) ) + ax1 ax1 + bx2 + cx3 (b + c) ( (cy 3 + by 2 ) / (c+ b) ) + ay1 ay1 + by 2 + cy 3 ∴ x = = , y = b+c+a a+b+c b+c+a a+b+c (b + c) ( (cy 3 + by 2 ) / (c+ b) ) + ay1 ay1 + by 2 + cy 3 y = b+c+a a+b+c  M a them a tics | 8.5 5.3 Ex-centres A The centre of the circle which touches the side BC and the extended portions of sides I3 I2 AB and AC is called the ex-centre of ∆ABC with respect to the vertex A. It is denoted by I1 and its co-ordinates are as follows: B C  −ax1 + bx2 + cx3 −ay1 + by 2 + cy 3  I1 =  ,   −a + b + c −a + b + c  I1 Similarly ex-centres of ∆ABC with respect to vertices B and C are denoted by I2 and I3, respectively, and Figure 8.8  ax – bx2 + cx3 ay1 − by 2 + cy 3  I2 =  1 ,  ,  a−b + c a−b + c   ax + bx2 − cx3 ay1 + by 2 − cy 3  I3 =  1 , .  a+b−c a+b−c   5.4 Circumcentre A(x1, y1) It is the point of intersection of perpendicular bisectors of the sides of the triangle. It is also the centre of a circle passing through the vertices of the triangle. If O is the O circumcentre of any ∆ABC, then, OA = OB = OC. B E C(x3, y3)  x sin2A + x2 sin2B + x3 sin2C y1 sin2A + y 2 sin2B + y 3 sin2C  Circumcentre:  1 ,  (x2, y2)  Σ sin2A Σ sin2A  Figure 8.9 Note: For a right-angled triangle, its circumcentre is the mid-point of hypotenuse. B A C Figure 8.10 A(x1, y1) 5.5 Orthocentre The point of intersection of altitudes of a triangle that can be obtained by solving the D F equation of any two altitudes is called Orthocentre. It is denoted by H H  x tanA + x2 tanB + x3 tanC y1 tanA + y 2 tanB + y 3 tanC  B E C Orthocentre:  1 ,  (x2, y2) (x3, y3)  Σ tanA Σ tanA  Figure 8.11 Note: In a right angle triangle, orthocentre is the point where right angle is formed. A Remarks: (a) In an equilateral triangle, centroid, incentre, orthocentre, circumcentre coincide. G H (b) Orthocentre, centroid, and circumcentre are always collinear. Centroid divides the O Orthocentre and circumcentre joining line in a 2: 1 ratio. B C Proof: H, G and O are collinear and ∆’s OGD & AGH are similar. D N But OD (distance of c.c. from BC) = R cos A Figure 8.12 8. 6 | Straight Line HA = distance of orthocentre from vertex A = 2R cos A AH AG HG ∴ =2= = ⇒ G divides line joining H and O in 2:1. OD GD GO (c) In an isosceles triangle centroid, orthocentre, incentre, circumcentre lie on the same line. 5.6 Nine-Point Circle Nine-point circle can be constructed for any given triangle, and is so named because it touches nine significant concyclic points throughout the triangle. These nine points are as follows: Mid-point of each side of the triangle Foot of each altitude Mid-point of the line segment from each vertex of the triangle to the orthocentre. L L Figure 8.13 MASTERJEE CONCEPTS The centroid, incentre, orthocentre and circumcentre coincide in an equilateral triangle. In an isosceles triangle, centroid, orthocentre, incentre and circumcentre lie on the same line. Orthocentre, centroid and circumcentre are always collinear, and centroid divides the line joining orthocentre and circumcentre in the ratio 2:1. Saurabh Gupta (JEE 2010, AIR 443) Illustration 4: If G be the centroid of the triangle ABC, prove that AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2).  (JEE MAIN) Sol: Distance formula of two points can be used to prove AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2). In triangle ABC, let B be the origin and BC the x-axis. Let A be (h, k) and a+h k  A(h, k) C be (a, 0). Then centroid G is  , .  3 3 LHS = AB2 + BC2 + CA2 = (h – 0)2 + (k – 0)2 + a2 + (h – a)2 + (k – 0)2 C (a, 0) = 2h2 + 2k2 + 2a2 – 2ah B Figure 8.14  M a them a tics | 8.7 ∴ RHS  a + h 2  k 2  a+h 2  k 2  a+h 2  k  = 3  − h +  − k  +  − 0 +  − 0 +  − a  +  − 0   3  3   3  3   3  3    = 1/3 [(a – 2h)2 + 4k2 + (a + h)2 + k2 + (h – 2a)2 + k2] = 2h2 + 2k2 + 2a2 – 2ah Hence, it is equal on both sides. C y 5.7 Area of a Triangle Let A, B, C be the vertices of the triangle having (x1, y1), (x2, y2) and (x3, y3) as their B respective co-ordinates. Draw AL, BM, CN perpendicular to the x-axis. A Then ∆ABC = trapezium ALNC + trapezium CNMB – trapezium ALMB 1 1 1 = (LA + NC) LN + (NC + MB) NM – (LA + MB) LM 2 2 2 x 1 1 1 O L N M = (y1 + y3) (x3 – x1) + (y3 + y2) (x2 – x3) – (y1 + y2) (x2 – x1) 2 2 2 Figure 8.15 1 = {x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)} 2 x1 y1 1 1 = x2 y2 1 2 x3 y3 1 If the area of the triangle formed by the points (x1, y1), (x2, y2) and (x3, y3) is zero, the points lie on a straight line. Using this, we can determine whether three point are in a straight line. i.e. the condition for (x1, y1), (x2, y2) and (x3, x1 y1 1 y3) to be collinear is that x2 y 2 1 = 0. x3 y3 1 1 For example, the area of the triangle formed by the points (1, 4), (3, –2) and (–3, 16) is {1(–2 –16) + 3} {(16 – 4) – 3 (4 + 2)} = 0. The three points lie on a straight line. 2 Illustration 5: The vertices of a triangle ABC are A(p2, –p), B(q2, q) and C(r2, –r). Find the area of the triangle.  (JEE MAIN) x1 y1 1 1 Sol: Area of the triangle formed by the points (x1, y1), (x2, y2) and (x3, y3) is x2 y 2 1. Substituting the given co-ordinates, we can obtain area of given triangle. 2 x y 1 3 3 p2 −p 1 p2 − q2 −(p + q) 0 p − q −1 0 1 2 1 2 2 1 D= q q 1 = q −r q+r 0 = (p + q) (q + r) q − r 1 0 2 2 2 2 r −r 1 r2 −r 1 r2 −r 1 1 1 = (p + q) (q + r) [(p – q) + (q – r)] = (p + q) (q + r) (p – r) 2 2 8. 8 | Straight Line Note: (a) If area of the triangle is zero, then the three points are collinear. (b) The area of a polygon with vertices A1(xi, yi), i = 1, …, n (vertices taken in anti-clockwise order) 1 [(x y – x2y1) + (x2y3 – x3y2) +…+(xny1 – x1yn)] 2 1 2 6. LOCUS Locus is a set of points which satisfies a given geometrical data. Thus, for example, locus of a point moving at a constant distance from a given point is a circle. Locus of a point which is equidistance from two fixed points is a perpendicular bisector of the line joining the two points. All the points in a locus can be represented by an equation. For example, (a) If the distance of the point (x, y) from (2, 3) is 4, then (x – 2)2 + (y – 3)2 = 42. i.e. x2 + y2 – 4x – 6y – 3 = 0. This equation will represent a circle with its centre at (2, 3) and radius 4. (b) If (x, y) be the point equidistant from the points (3, 4) and (2, 1), then (x – 3)2 + (y – 4)2 = (x – 2)2 + (y – 1)2 i.e. x + 3y = 10. From the geometrical constraint, which governs the motion, we can find a relation (locus) between the co- ordinates of the moving point in any of its positions. Equation of locus is therefore merely on equation relating the x and y co-ordinates of every point on the locus. Steps to find locus (i) Assume the co-ordinates of point for which locus is to be determined as (h, k). (ii) Apply the given geometrical conditions. (iii) Transform the geometrical conditions into algebraic equation and simplify. (iv) Eliminate variables (if any). (v) Replace h → x and k → y to get the equation of locus. Note: Locus should not contain any other variables except x and y. The algebraic relation between x and y satisfied by the co-ordinates at every point on the curve and not off the curve is called the equation of curve. y Illustration 6: Find the equation of locus of a point which moves so that its distance P(x, y) (0, 1) from the point (0, 1) is twice the distance from x-axis. (JEE MAIN) N Sol: Here we can obtain the equation of locus of given point by using given condition and distance formula of two points. x Let the co-ordinates of such a point be (x, y). Draw PM ⊥ to x-axis. x’ 0 M y’ Hence, PM = y Figure 8.16 PN = 2PM (given)  M a them a tics | 8.9 i.e. (x – 0)2 + (y – 1)2 = 4y2 i.e. x2 – 3y2 – 2y + 1 = 0. Illustration 7: Locus of the centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, – b cos t) and (1, 0), where t is a parameter is  (JEE MAIN) (A) (3x – 1)2 + (3y)2 = a2 + b2 (B) (3x + 1)2 + (3y)2 = a2 + b2 (C) (3x + 1)2 + (3y)2 = a2 – b2 (D) (3x – 1)2 + (3y)2 = a2 – b2 Sol: The centroid (h, k) of a triangle formed by points (x1, y1), (x2, y2) and (x3, y3) will be x1 + x2 + x3 y1 + y 2 + y 3 h = and k. 3 3 (A) If (h, k) is the centroid, then acos t + bsint + 1 asint − bcos t + 0 h= ,k= ⇒ (3h – 1)2 + (3k)2 = (a cos t + b sin t)2 + (a sint t – b cos t)2 = a2 + b2 3 3 ∴ Locus of (h, k) is (3x – 1)2 + (3y)2 = a2 + b2 7. STRAIGHT LINE Definition: It is defined as the locus of a point such that any two points of this locus have a Y constant inclination (gradient). Inclination: If a straight line intersects the x-axis, the inclination of the line is defined  as the measure of the smallest non-negative angle which the line makes with the X’ positive direction of the x-axis 180-   π Slope (or gradient): If the inclination of a line (i.e. non-vertical line) is θ and  θ ≠  , then the slope of a line is defined to be tan θ and is denoted by m.  2 Y’ ∴ m = tanq Figure 8.17 (a) Slope of x-axis is zero. (b) Slope of y-axis is not defined. 7.1 Slope (x2, y2)Q l Let P(x1, y1) and Q(x2, y2) be two points on a line, then slope will be y2 - y1 (x1, y1)P MQ M m = tan= θ x2 - x 1 MP y 2 − y1  m = tan= θ x2 − x1 O x y 1 Figure 8.18 (a) Line passing through two points (x1, y1) and (x2, y2) is x1 y1 1 = 0 x2 y2 1 Note: Above-mentioned matrix form is a condition for three points to be collinear. (x1, y1) (x2, y2) (b) Equation of the median through A(x1, y1) is Figure 8.19 8. 1 0 | Straight Line A(x1, y1) P(x, y) B C M (x3, y3) (x2, y2) ( x2 +x3 y2 +y3 2 , 2 ) Figure 8.20 x y 1 x y 1 x y 1 x1 y1 1 = 0 or x1 y1 1 + x1 y1 1 = 0 x 2 + x3 y 2 + y3 x2 y2 1 x3 y3 1 1 2 2 (c) Equation of internal and external angle bisectors of A are A(x1, y1) x y 1 x y 1 b x1 y1 1 ± c x1 y1 1 = 0 P(x, y) x2 y2 1 x3 y3 1 P(x, y) b (x2, y2)B b E D C(x3, y3) ( bx2 + cx3 by2 + cy3 b+c , b+c ) ( bx2 - cx3 by2 - cy3 b-c , b-c ) Figure 8.21 7.2 Angle between Two Lines Two lines intersecting each other make two angles between them, one acute and the other obtuse. Figure 8.22 shows lines L1 and L2 intersecting each other, acute angle θ and obtuse angle φ. Let line L1 makes angle q1 with x-axis and L2 makes q2. y L2 L1    1 -2 A x O B Figure 8.22  M a them a ti cs | 8.11 Therefore slope of L1 is m1= tan q1 Slope of L2 is m2 = tanq2 Now in ∆ABC, q1 + π – q2 + θ = π θ = θ2 – θ1 tan θ2 − tan θ1 m2 − m1 tanθ = tan (q2 – q1) ⇒ tanθ = ⇒ tan θ = , this gives the acute angle between lines. 1 + tan θ2. tan θ1 1 + m1m2 Note: (i) If m1 = m2, then θ = 0º, i.e. lines are parallel or coincident. (ii) If m1m2 = −1, then θ = 90 º , i.e. lines are perpendicular to each other. π 1 Illustration 8: If the angle between two lines is and slope of one of the lines is. Find the slope of the other.  4 2 (JEE MAIN) m2 – m1 Sol: We know that, tan θ = , where m1 and m2 are the slope of lines and θ is the angle between them. 1+m1m2 1 π π m – (1 / 2) m – (1 / 2) Let m1 = , m2 = m and θ = So, tan = ⇒ 1 =± ⇒ m = 3 or –(1/3) 2 4 4 1+(1 / 2)m 1+(1 / 2)m Illustration 9: Line through the point (–2, 6) and (4, 8) is perpendicular to the line through the point (8, 12) and (x, 24). Find the value of x.  (JEE MAIN) Sol: Given two lines are perpendicular to each other. Therefore, product of their slope will be -1. 8–6 2 1 Slope of the line through the points (–2, 6) and (4, 8) is m1= = = 4 – (–2) 6 3 24 – 12 12 Slope of the line through the points (8, 12) and (x, 24) is m2 = = x–8 x–8 Since two lines are perpendicular m1m2 = – 1 1 12 ⇒ × = –1 ⇒x=4 3 x–8 7.3 Collinearity C B A Figure 8.23 If three points A, B, C are collinear, then Slope of AB = Slope of BC = slope of AC 8. 1 2 | Straight Line MASTERJEE CONCEPTS Collinearity of three given points: Three given points A, B, C are collinear if any one of the following conditions is satisfied. Area of triangle ABC is zero. Slope of AB = Slope of BC = Slope of AC. AC = AB + BC. Find the equation of the line passing through two given points, if the third point satisfies the equation of the line, then three points are collinear. 1 If any one line is parallel to y-axis, then the angle between two straight lines is given by tan θ = ± , where m is the slope of other straight line. m A line of gradient m is equally inclined with the two lines of gradient m1 and m2. m1 – m m2 – m Then =-. 1+m1m 1+m2m Aman Gour (JEE 2012, AIR 230) 7.4 Equation of a line (a) Point slope form: Suppose P0(x0, y0) is a fixed point on a non-vertical line L whose P(x, y) slope is m. Let P(x, y) be an arbitrary point on L. Then by definition, the slope of L y – y0 P0(x0, y0) is given by m = ⇒ y – y 0 = m(x – x0 ) x – x0 This is called point slope form of a line. Figure 8.24 (b) Two point form: Let line L passes through two given points P1(x1, y1) and P2(x2, y2). Let P(x, y) be a point on the line. So slope P1P = slope P1P2 y – y1 y 2 – y1 ⇒ = P2(x2, y2) x – x1 x2 – x1 P(x, y) y 2 – y1 ⇒ y – y1 = (x – x1 ) P1(x1, y1) x2 – x1 This is called two-point form of the line. Figure 8.25 (c) Slope intercept from: Case-I: If slope of line is m and makes y-intercept c, then equation is (y – c) = m (x – 0) ⇒ y = mx + c Case-II: If slope of line is m and makes x-intercept d, then equation is y = m(x – d) These equations are called slope intercept form. (0, c) (d, 0) Figure 8.26  Mathematics | 8.13 (d) Intercept form: Suppose a line L makes intercept a on x-axis and intercept b on L y-axis, i.e. the line meets x-axis at (a, 0) and y-axis (0, b). (0, b) b–0 So, y – 0 = (x – a) b 0–a x y i.e. + = 1. This is called intercept form of the line. a b a (a, 0) Figure 8.27 (e) Normal form: If P is perpendicular distance from origin to the line AB and makes angle α with x-axis, then equation of the line is x cos α + y sin α = P OM y Proof: cos α = OM = OL cos α = x cos α OL B In DPNL, PN sin α = PL Q PN = PL sin α = y sin α M P(x,y) 90- MQ = PN = y sin α  N  x O x A Now P = OQ = OM + MQ = x cos α + y sin α L Figure 8.28 So x cos α + y sin α = P This is called normal form of the line. (f) Parametric form or distance form: The equation of the line passing through (x1, y1) and making an angle θ with the positive x-axis is x – x1 y – y1 = = r , where ‘r’ is the signed value. cos θ sin θ Hence, the co-ordinate of any point at a distance r on this line is x = x1 + r cos θ y = y1 + r sin θ MASTERJEE CONCEPTS Point of intersection of two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is given by  b1 b2 c1 c2     b1c2 – b2c1 c1a2 – c2a1   c1 c2 a1 a2  ( x', y')=  a b – a b , a b – a b  =  a1 a2 , a1 a2   1 2 2 1 1 1 2 1    b1 b2 b1 b2   Saurabh Gupta (JEE 2010, AIR 443) 8. 1 4 | Straight Line Illustration 10: A straight line is drawn through the point P(2, 3) and is inclined at an angle of 30° with positive x-axis. Find the co-ordinate of two points on it at a distance 4 from P on either side of P.  (JEE MAIN) x – x1 y – y1 Sol: By using formula = = r , we can obtain co-ordinates of point. cos θ sinθ The equation of line x – x1 y – y1 x–2 y –3 = = ±r ⇒ = = ± 4 ⇒ x = 2 ± 2 3, y = 3 ± 2 cos θ sinθ cos 30° sin 30° So, co-ordinate of two points are 2± 2 3 ,3± 2 ( ) Illustration 11: If two vertices of a triangle are (–2, 3) and (5, –1). Orthocentre of the triangle lies at the origin and centroid on the line x + y = 7, then the third vertex lies at  (JEE MAIN) (A) (7, 4) (B) (8, 14) (C) (12, 21) (D) None of these Sol: (D) The line passing through the third vertex and orthocentre must be perpendicular to line through (–2, 3) and (5, –1). Therefore, product of their slope will be -1. Given the two vertices B(–2, 3) and C(5, – 1); let H(0, 0) be the orthocentre; A(h, k) the third vertex. Then, the slope of the line through A and H is k/h, while the line through B and C has the slope (–1 –3)/(5 + 2) = – 4/7. By the property of the orthocentre, these two lines must be perpendicular,  k  4  k 7 So we have    −  = – 1 ⇒ =  … (i) h   7  h 4 5 − 2 + h −1 + 3 + k Also + = 7 ⇒ h + k = 16 … (ii) 3 3 Which is not satisfied by the points given in (A), (B) or (C). Illustration 12: In what direction should a line be drawn passing through point (1, 2) so that its intersection point 6 with line x + y = 4 is at a distance of units.  (JEE ADVANCED) 3 Sol: By using x = x1 + r cos θ and y = y1 + r sin θ, we can obtain the required angle. For co-ordinates of B 6 6 6 Substitute r= ∴ x = 1+ cos θ & = y 2+ sin θ B 3 3 3 A(1,2) Substituting in x + y =4 6  3 6 6 3 ⇒ 1+ cos θ + 2+ sin θ = 4 ∴ (cos θ + sin θ) = 3 3 6 1 1 3 1 ∴ cos θ + sin θ = (Multiple by ) 2 2 2 2 Figure 8.29 sin (45° + θ) = sin 60° ∴ θ = 15° or, sin (45° + θ)= sin 120° ∴ θ = 75°  Mathematics | 8.15 Illustration 13: If sum of the distances of the points from two perpendicular lines in a plane is 1, then find its locus. (JEE ADVANCED) Sol: If (h, k) be any point on the locus, then |h| + |k| = 1 Y Let the two perpendicular lines be taken as the co-ordinate axes. ⇒ locus of (h, k) is |x| + |y| = 1 (0, 1) This consists of four line segments which enclose a square as -1 x y= + y x- shown in figure. =1 X  (-1, 0) O (1, 0) x =1 + y y x- =1 (0, -1) Figure 8.30 Illustration 14: If the circumcentre of a triangle lies at the origin and the centroid is the mid-point of the line joining the points (a2 + 1, a2 + 1) and (2a, –2a), then the orthocentre lies on the line.  (JEE ADVANCED) (A) y = (a2 + 1)x (B) y = 2ax (C) x + y = 0 (D) (a – 1)2 x – (a + 1)2 y = 0 Sol: (D) We know from geometry that the circumcentre, centroid and orthocentre of a triangle lie on a line. So the  (a + 1)2 (a − 1)2  orthocentre of the triangle lies on the line joining the circumcentre (0, 0) and the centroid  ,   2 2  2 2  (a + 1) (a − 1) y= x or (a – 1)2 x – (a + 1)2 y = 0. 2 2 MASTERJEE CONCEPTS Equation of parallel and perpendicular lines: Equation of a line which is parallel to ax + by + c = 0 is ax + by + k = 0. Equation of a line which is perpendicular to ax + by + c = 0 is bx – ay + k = 0. If y = m1x + c1, y = m1x + c2, y = m2x + d1 and y = m2x + d2 are sides of a parallelogram then its (c1 – c2 )(d1 – d2 ) area is. m1 – m2 x y The equation of a line whose mid-point is (x1, y1) in between the axes is + =2. x1 y1 c2 Area of the triangle made by the line ax + by +c = 0 with the co-ordinate axes is. 2 | ab | A line passing through (x1, y1) and if the intercept between the axes is divided in the ratio m:n at this nx my point then the equation is + = m+n. x1 y1 The equation of a straight line which makes a triangle with the co-ordinates axes whose centroid is x y (x1, y1) is + =1. 3x1 3y1 B Rajiv Reddy (JEE 2012, AIR 11) 8. 1 6 | Straight Line 7.5 Foot of the Perpendicular A(x1, y1) ax+by+c=0 The foot of the perpendicular (h, k) from (x1, y1) to the line ax + by + c = 0 is given by h – x1 k – y1 –(ax1 +by1 + c) = =. a b a2 +b2 Hence, the co-ordinates of the foot of perpendicular is  b2 x – aby – ac a2 y – abx – bc   1 1 , 1 1 .  a2 +b 2 a2 +b 2  B(h, k)   Figure 8.31 The image of a point with respect to the line mirror: The image of A(x1, y1) with respect to the line mirror ax + h – x1 k – y1 –2(ax1 +bx1 + c) by + c = 0, B (h, k) is given by = = a b a2 +b2. Special Cases (a) Image of the point P(x1, y1) with respect to x-axis is (x1, – y1). (b) Image of the point P(x1, y1) with respect to y-axis is (– x1, y1). (c) Image of the point P(x1, y1) with respect to the line mirror y = x is Q(y1, x1). (d) Image of the point P(x, y) with respect to the origin is the point (-x, -y). Illustration 15: Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the line makes an angle of 30° with the positive direction of the x-axis.  (JEE MAIN) Sol: By using xcosα + ysinα = P, we can solve this problem. Here α = 30° and P = 3. 3 y So equation is x cos 30° + y sin 30° = 3 x + =3 ⇒ 3x + y = 6 2 2 Position of a point w.r.t. a line Let the equation of the given line be ax + by + c = 0 and let the co-ordinates of the two given points be P(x1, y1) and Q(x2, y2). Let R1 be a point on the line. m –ax1 – by1 – c The co-ordinates of R1 which divides the line joining P and Q in the ratio m:n are = n ax2 +by 2 + c. Thus, the two points (x1, y1) and (x2, y2) are on the same (or opposite) sides of the straight line ax + by + c = 0 m whether Point R1 divides internally or externally or sign of. n Note: ⇒ A point (x1, y1) will lie on the side of the origin relative to a line ax + by + c = 0, if ax1 +by1 + c and c have the same sign. ⇒ A point (x1, y1) will lie on the opposite side of the origin relative to the line ax + by + c = 0, if ax1 + by1 + c and c have the opposite sign. Illustration 16: For what values of the parameter α does the point M (α, α + 1) lies within the triangle ABC of vertices A(0, 3), B(– 2, 0) and C(6, 1).  (JEE ADVANCED) Sol:Here, the point M will be inside the triangle if and only if |Area ∆MBC| + |Area ∆MCA| + |Area ∆MAB| = |Area ∆ABC|. And each individual area must be non-zero.  M a them a ti cs | 8.17 α α +1 1 1 1 Area MBC = –2 0 1 = 7α + 6 2 2 6 1 1 α α +1 1 1 1 Area MCA = 6 1 1 = –8α +12 2 2 0 3 1 α α +1 1 1 1 Area MAB = 0 3 1 = α+4 2 2 –2 0 1 Figure 8.32 0 3 1 1 1 Area ABC = –2 0 1 =.22 2 2 6 1 1 6 3 The above equation has critical points – 4, – and. 7 2 For α ≤ –4, the equation is –7α – 6 – 8α + 12 – α – 4 =22 5 5 ⇒ α=– which is not a solution, since – > –4 4 4  6 6 For a∈  –4, –  , then equation is – 7α – 6 – 8α + 12 + α + 4 = 22 ⇒ α = –  7 7 6 which is solution of equation but area MBC = 0 ⇒ M lies on BC ⇒ α = – is not the desired value. 7  6 3 For a∈  – ,  , the equation is 7α + 68α + 12 + α + 4 = 22.  7 2  6 3 ⇒ All α in the interval  – ,  satisfy the equation.  7 2 3  3 Finally over  , ∞  , we get α = implies area MCA become zero.  2  2  6 3 ⇒ The desired values of α lie in the interval  – , .  7 2 7.6 Length of the Perpendicular P(x1,y1) The perpendicular distance ‘p’ of a point P(x1, y1) from the line p | ax1 +by1 + c | ax + by + c = 0 is p= ax+by+c a2 +b2 M Figure 8.33 (a) Distance between parallel lines: The distance between the parallel lines a x+ by + c1 = 0 and ax + by + c2 = 0 is c1 – c2 a2 +b2 y=mx+c (  (b) Lines making angle α with given line: The equations of the two straight lines ( passing through P(x’, y’) and making an angle α with the line y = mx + c P(x’,y’) (where m = tan θ) are Figure 8.34 8. 1 8 | Straight Line y – y’ = tan (θ + α) (x – x’) π Note: If θ + α or θ – α is an odd multiple of , the corresponding line has equation x = x’. 2 (c) Concurrency of lines: Lines aix + biy + c1 = 0, where i = 1, 2, 3 are concurrent if they meet at a point. The a1 b1 c1 condition for concurrency is a2 b2 c2 = 0 a3 b2 c3 Illustration 17: The equation of the two tangents to the circle are 3x – 4y + 10 = 0 and 6x – 8y + 30 = 0. Find diameter of the circle.  (JEE MAIN) c1 – c2 Sol: By using formula of distance between two parallel line, i.e. , we can find the a2 +b2 diameter of given circle. d These are two parallel lines 3x – 4y + 10 = 0.....(i) Figure 8.35 6x – 8y + 30 = 0.....(ii) 15 − 10 Dividing second equation by 2 gives 3x – 4y + 15 = 0; ∴ d = =1 32 + 42 MASTERJEE CONCEPTS (a) A triangle is isosceles if any two of its median are equal. (b) Triangle having integral co-ordinates can never be equilateral. (c) If arx + bry + cr = 0 (r = 1, 2, 3) are the sides of a triangle then the area of the triangle is given 2 a1 b1 c1 1 by a2 b2 c2 where C1, C2 and C3 are the cofactors of c1, c2 and c3 in the determinant. 2C1C2C3 a3 b3 c3 (d) Area of parallelogram: (i) Whose sides are a and b and angle between them is θ is given by ab sin q. Area of ABCD = ab.sinθ a D C b b  A a B Figure 8.36 (ii) Whose length of perpendicular from one vertices to the opposite sides are p1 and p2 and angle PP between sides is θ is given by Area = 1 2 Sin θ p2 p1 θ Figure 8.37 Krishan Mittal (JEE 2012, AIR 199)  M a them a ti cs | 8.19 8. FAMILY OF LINES Consider two intersecting lines L1: a1x + b1y + c1 = 0 and L2: a2x + b2y + c2 = 0, then Type-1: The equation of the family of lines passing through the intersection of the lines L1 + λL2 = 0 ⇒ (a1x +b1y + c1) + λ(a2x +b2y + c2) = 0 where λ is a parameter. Type-2: Converse, L1 + λL2 = 0 is a line which passes through a fixed point, where L1 = 0 and L2=0 are fixed lines and the fixed point is the intersection of L1 and L2. Type-3: Equation of AC ≡ u2u3 – u1u4 = 0 and BD ≡ u3u4 – u1u2= 0 u1a1x+b1y+c1=0 B C u2a2x+b2y+d2=0 u4a2x+b2y+d1=0 A D u3a1x+b1y+c2=0 Figure 8.38 Note that second degree terms cancel and the equation u2u3 – u1u4 = 0 is satisfied by the co-ordinate points B and D. Illustration 18: If a, b, c are in A.P., then prove that the variable line ax + by + c = 0 passes through a fixed point.  (JEE MAIN) Sol: By using given condition we can reduce ax + by + c = 0 to as L1 + λL2 = 0. Hence we can obtain co-ordinate of fixed point by taking L1 = 0 and L2=0. 2b =a + c ⇒ c = 2b – a ⇒ ax + by + 2b – a = 0 ∴ a (x – 1) + b(y + 2) = 0 This is of the form L1 + λL2 = 0, where b/a = l ∴ Co-ordinates of fixed point is (1, – 2). 9. ANGULAR BISECTOR 9.1 Bisectors of the Angle Between Two Lines (a) Equations of the bisectors of angle between the lines ax + by + c = 0 and a1x + b1y + c1 = 0 are ax +by + c a1 x +b1 y + c1 =± (ab1 ≠ a1b) a2 +b2 a12 +b12 (b) To discriminate between the bisectors of the angle containing the origin and that of angle not containing the origin, rewrite the equations, ax + by + c = 0 and a’x+ b’y + c’ = 0 such that the terms c, c’ are positive, ax +by + c a'x +b'y + c' then =+ gives the equation of the bisector of the angle containing origin and 2 2 a +b a'2 +b'2 ax +by + c a'x +b'y + c' =– gives the equation of the bisector of the angle not containing origin. 2 2 a +b a'2 +b'2 (c) Acute angle bisector and obtuse angle bisector can be differentiated from the following methods: Let two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 intersect such that constant terms are positive. 8. 2 0 | Straight Line If a1a2 + b1b2 < 0, then the angle between the lines that contain the origin is acute and the equation for a x +b1 y + c a x +b2 y + c2 ax1 +by1 + c a x +b2 y + c2 the acute angle bisector is 1 =+ 2. Therefore =– 2 is the 2 2 2 2 2 2 a +b a2 +b2 a +b a22 +b22 equation of other bisector. If, however, a1a2 + b1b2 > 0, then the angle between the lines containing the origin a1 x +b1 y + c1 a2 x +b2 y + c2 is obtuse and the equation of the bisector of the obtuse angle is =+ ; therefore a1 x +b1 y + c1 a2 x +b2 y + c2 a12 +b12 a22 +b22 =– is acute angle bisector. a12 +b12 a22 +b22 (d) Few more methods of identifying an acute and obtuse angle bisectors are as follows: Let L1 = 0 and L2 = 0 are the given lines and u1 = 0 and u2 = 0 are the bisectors L1=0 between L1 = 0 and L2 = 0. Take a point P on any one of the lines L1 = 0 or L2 = 0 and draw a perpendicular on u1 = 0 and u2 = 0 as shown. If P p q |p| < |q| ⇒ u1 is the acute angle bisector. L2=0 |p| > |q| ⇒ u1 is the obtuse angle bisector. |p| = |q| ⇒ the lines L1 and L2 are perpendicular. Note: The straight lines passing through P(x1, y1) and equally inclined with the lines u2=0 u1=0 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are those which are parallel to the bisectors between lines and passing through the point P. Figure 8.39 MASTERJEE CONCEPTS (a) Algorithm to find the bisector of the angle containing the origin: Let the equations of the two lines be a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. The following methods are used to find the bisector of the angle containing the origin: Step I: In the equations of two lines, check if the constant terms c1 and c2 are positive. If the terms are negative, then make them positive by multiply both the sides of the equation by –1. Step II: Obtain the bisector corresponding to the positive sign, i.e. a1 x +b1 y + c1 a2 x +b2 y + c2 = a12 +b12 a22 +b22 L1 Acute bisector L2 Obtuse bisector Figure: 8.40 This is the required bisector of the angle containing the origin, i.e. the bisectors of the angle between the lines which contain the origin within it. (b) Method to find acute angle bisector and obtuse angle bisector (i) Make the constant term positive by multiplying the equation by –1. (ii) Now determine the sign of the expression a1a2 + b1b2.  Mathematics | 8.21 MASTERJEE CONCEPTS (iii)If a1a2 + b1b2 > 0, then the bisector corresponding to ‘+ve’ and ‘–ve’ signs give the obtuse and acute angle bisectors, respectively, between the lines. (iv)If a1a2 +b1b2 < 0, then the bisector corresponding to ‘+ve’ and ‘–ve’ signs give the acute and obtuse angle bisectors, respectively. Both the bisectors are perpendicular to each other. If a1a2 + b1b2 > 0, then the origin lies in the obtuse angle and if a1a2 + b1b2 < 0, then the origin lies in the acute angle. T P Varun (JEE 2012, AIR 64) MASTERJEE CONCEPTS Incentre divides the angle bisectors in the ratios (b + c):a, (c + a):b and (a + b):c. Angle bisector divides the opposite sides in the ratio of remaining sides. Figure: 8.41 BD AB c = = DC AC b Aishwarya Karnawat (JEE 2012, AIR 839) 9.2 Bisectors in Case of Triangle Two possible models are as follows: Case-I: When vertices of a triangle are known, compute the sides of the A(x1,y1) triangle and the incentre. All the internal bisectors can be known, using the co-ordinates of incentre and vertices of triangle. Note: If the triangle is isosceles/equilateral, then one can easily get the c b incentre. Case-II: When the equations of the sides are given,compute tan A, tan B, (x2,y2)B C(x3,y3) tan C by arranging the lines in descending order of their slope. Compute the a acute/obtuse angle bisectors as the case may be. Plot the lines approximately Figure 8.42 and bisectors containing or not containing the origin. 8. 2 2 | Straight Line Illustration 19: The line x + y =a meets the x- and y-axes at A and B, respectively. A triangle AMN is inscribed in the triangle OAB, O being the origin, with right angle at N. M and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to. (JEE ADVANCED) (A) 3 (B) 1/3 (C) 2 (D) 1/2 Sol: (A) Here simply by using the formula of area of triangle, Y 1 B(0, a) i.e. {x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)} , we can solve the problem. 2 N AN  a λa  Let = λ. Then the co-ordinates of N are  , , BN  1 + λ 1 +λ M where (a, 0) and (0, a) are the co-ordinates of A and B, respectively. Now equation of MN perpendicular to AB is X O A(a, 0) λa a 1–λ  λ –1  y– =x– or x – y = a So the co-ordinates of M are  0, a 1+ λ 1+ λ 1+λ  λ +1  Figure 8.43 Therefore, area of the triangle AMN is 1   –a  1 – λ 2 λa2 = a    + a  = 2   λ + 1  (1 + λ ) 2  (1 + λ )2 Also area of the triangle OAB = a2/2. λa2 3 1 2 So that according to the given condition: =. a ⇒ 3l2 – 10λ + 3 = 0 ; ⇒ λ = 3 or λ = 1/3. (1 + λ ) 2 8 2 For λ = 1/3, M lies outside the segment OB and hence the required value of λ is 3. 10. PAIR OF STRAIGHT LINES 10.1 Pair of Straight Lines Through Origin (a) A homogeneous equation of degree two of the type ax2 + 2hxy + by2 = 0 always represents a pair of straight lines passing through the origin and if (i) h2 > ab ⇒ lines are real and distinct. (ii) h2 = ab ⇒ lines are coincident. (iii) h2 < ab ⇒ lines are imaginary with real point of intersection, i.e. (0, 0) (b) If y = m1x and y = m2x be the two equations represented by ax2 + 2hxy + by2 = 0, then 2h a m1 + m2 = – and m1m2 = b b Angle between two straight lines: (c) If θ is the acute angle between the pair of straight lines represents by ax2 + 2hxy + b, then 2 h2 – ab tan θ = a+b The condition that these lines are: (i) At right angles to each other if a + b = 0, i.e. sum of coefficients of x2 and y2 is zero.  Mathematics | 8.23 (ii) Coincident if h2 = ab and (ax2 + 2hxy + by2) is a perfect square of ( ax + by)2. (iii) Equally inclined to the axis of x if h = 0, i.e. coefficient of xy =0. Combined equation of angle bisectors passing through origin: The combined equation of the bisectors of the angles between the lines ax2 + 2hxy + by 2 = 0 (a pair of straight lines passing through origin) is given by x2 – y 2 xy =. a–b h 10.2 General Equation for Pair of Straight Lines (a) ax2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represents a pair of straight lines if: a h g 2 2 2 abc + 2fgh – af – bg – ch = 0 , if h b f = 0 g f c (b) The slope of the two lines represented by a general equation is the same as that between the two lines represented by only its homogeneous part. 10.3 Homogenisation The equation of the two lines joining the origin to the points of intersection of the line lx + my +n = 0 and the curve ax2 + 2hxy + by 2 + 2gx + 2fy + c = 0 is obtained by homogenising the equation of the curve using the equation of the line. The combined equation of pair of straight lines joining origin to the points y of intersection of the line given by lx + my + n = 0  …. (i) The second degree curve: lx+my+n=0 ax2 + 2hxy + by 2 + 2gx + 2fy + c = 0  …. (ii) Using equation (i) and (ii) x O 2  lx + my   lx + my   lx + my  ax2 + 2hxy + by 2 + 2gx   + 2fy   + c  =0  …. (iii) Figure 8.44  –n   –n   –n   lx + my  Obtained by homogenizing (ii) with the help of (i), by writing (i) in the form:   = 1.  –n  MASTERJEE CONCEPTS Through a point A on the x-axis, a straight line is drawn parallel to y-axis so as to meet the pair of straight lines. ax2 + 2hxy + by 2 = 0 in B and C. If AB = BC, then 8h2 = 9ab. Krishan Mittal (Jee 2012, Air 199) 8. 2 4 | Straight Line Illustration 20: The orthocentre of the triangle formed by the lines xy = 0 and x + y = 1 is  (JEE MAIN) (A) (1/2, 1/2) (B) (1/3, 1/3) (C) (0, 0) (D) (1/4, 1/4) Sol: (C) Here the three lines are x = 0, y = 0 and x + y = 1. Since the triangle formed by the line x = 0, y = 0 and x + y = 1 is right angled, the orthocentre lies at the vertex (0, 0), the point of intersection of the perpendicular lines x = 0 and y = 0. Illustration 21: If θ is an angle between the lines given by the equation 6x2 + 5xy – 4y 2 + 7x + 13y – 3 = 0 then equation of the line passing through the point of intersection of these lines and making an angle θ with the positive x-axis is  (JEE ADVANCED) (A) 2x + 11y + 13 = 0 (B) 11x – 2y + 13 = 0 (C) 2x – 11y + 2 = 0 (D) 11x + 2y – 11 = 0 Sol: (B) By taking the term y constant and using the formula of roots of quadratic equation, we can get the 2 h2 – ab equation of two lines represented by the given equation and then by using tan θ = , we will get the required result. a+b Writing the given equation as a quadratic in x, we have –(5y + 7) ± (5y + 7)2 + 24(4y 2 – 13y + 3) 6x2 + (5y + 7) x – (4y 2 – 13y + 3) = 0⇒ x= 12 2 –(5y + 7) ± 121y – 242y + 121 –(5y + 7) ± 11(y – 1) 6y – 18 –16y + 4 = = = or 12 12 12 12 ⇒ 2x – y + 3 = 0 and 3x + 4y – 1 = 0 are the two lines represented by the given equation and the point of intersection is (– 1, 1), obtained by solving these equations. 2 h2 – ab 2 (5 / 2)2 – 6(–4) 121 11 Also tan θ = , where a = 6, b = – 4, h = 5/2 = = = a+b 6–4 4 2 11 So the equation of the required line is y= –1 (x + 1) ⇒ 11x – 2y + 13 = 0 2 Illustration 22: If the equation of the pair of straight lines passing through the point (1, 1), and making an angle θ with the positive direction of x-axis and the other making the same angle with the positive direction of y-axis is x2 – (a + 2)xy + y 2 + a(x + y – 1) = 0 , a ≠ – 2, then the value of sin 2θ is  (JEE ADVANCED) 2 2 (A) a – 2 (B) a + 2 (C) (D) (a + 2) a Sol: (C) As both line passes through (1, 1) and one line makes angle θ with x-axis and other line with y–axis, slopes of line are tan θ and cot θ Equations of the given lines are y – 1 = tan θ (x – 1) and y – 1 = cot θ (x – 1) So, their combined equation is [(y – 1) – tan θ (x – 1)] [(y – 1) – cot θ (x – 1) ] = 0 ⇒ (y – 1)2 – (tan θ + cot θ) (x – 1) (y – 1) + (x – 1)2 = 0 ⇒ x2 – (tan θ + cot θ) xy + y2 + (tan θ + cot θ – 2) (x + y – 1) = 0 Comparing with the given equation we get tan θ + cot θ = a + 2 1 2 ⇒ = a + 2 ⇒ sin2θ = sin θ cos θ a+2  Mathematics | 8.25 Illustration 23: If two of the lines represented by x 4 + x3 y + cx2 y 2 – xy 3 + y 4 = 0 bisect the angle between the other two, then the value c is (JEE ADVANCED) (A) 0 (B) –1 (C) 1 (D) – 6 Sol: (D) As the product of the slopes of the four lines represented by the given equation is 1 and a pair of line represents the bisectors of the angles between the other two, the product of the slopes of each pair is –1. So let the equation of one pair be ax2 + 2hxy – ay2 = 0. x2 – y 2 xy The equation of its bisectors is =. 2a h By hypothesis x 4 + x3 y + cx2 y 2 – xy 3 + y 4 ≡ (ax2 + 2hxy – ay2) (hx2 – 2axy – hy2) = ah(x4 + y4) + 2(h2 – a2) (x3y – xy3) – 6ahx2y2 Comparing the respective coefficients, we get ah =1 and c = –6ah = –6 11. TRANSLATION AND ROTATION OF AXES Y Y’ P 11.1 Translation of Axes O’ X’ Let OX and OY be the original axes, and let the new axes, parallel to original axes, be O’X’ and O’Y’. Let the co-ordinates of the new origin O’ referred to the original axes be (h, k). If X the point P has co-ordinates (x, y) and (x’, y’) with respect to original and new axes, O respectively, then x = x’ + h; y = y’ + k Figure 8.45 11.2 Rotation of Axes Let OX and OY be the original system of axes and let OX’ and OY’ be the new system Y of axes and angle XOX’ = θ (the angle through which the axes are turned). If the point P Y’ X’ P has co-ordinates (x, y) and (x’, y’) with respect to original and new axes, respectively, then x = x’ cos θ – y’ sin θ and y = x’ sin θ + y’ cos θ  X O in matrix form it is as follows: Figure 8.46  x  cos θ − sin θ   x'   =    y   sin θ cos θ   y ' MASTERJEE CONCEPTS If origin is shifted to point (α, β) , then new equation of curve can be obtained by putting x + α in place of x and y + β in placed of y. Vaibhav Krishnan (JEE 2009, AIR 22) Illustration 24: The line L has intercepts a and b on the co-ordinate axes. The co-ordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then 1 1 1 1 – + – is equal to  (JEE MAIN) 2 2 2 a p b q2 (A) –1 (B) 0 (C) 1 (D) None of these 8. 2 6 | Straight Line Sol: (B) By using intercept form of equation of line, we will get equation of line before and after rotation. As their perpendicular length from the origin does not change, by using distance formula the result can be obtained. x y X Y Equation of the line L in the two co-ordinate system is + = 1, + = 1 Where (x, y) are the new co-ordinates a b p q of a point (x, y) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed: 1 1 1 1 1 1 1 1 1 1 = ⇒ + = + ⇒ – + – =0 2 2 2 2 2 2 2 (1 / a2 ) + (1 / b2 ) (1 / p2 ) + (1 / q2 ) a b p q a p b q2 Illustration 25: Let 0 < α < π/2 be a fixed angle. If P = (cos θ, sin θ) and Q = (cos (α – θ), (sin (α – θ)). Then Q is obtained from P by  (JEE ADVANCED) (A) Clockwise rotation around origin through an angle α. (B) Anti-clockwise rotation around origin through an angle α. (C) Reflection in the line through the origin with slope tan α. (D) Reflection in the line through the origin with slope tan α/2. Sol: As we know angle decreases during clockwise rotation and increases during anticlockwise rotation. D Clockwise rotation of P through an angle α takes it to the point (cos (θ – α), sin(θ – α)) and anticlockwise takes it to (cos (α + θ), sin (α + θ)) sin θ – sin(α – θ) 2cos(α / 2)sin(θ – α / 2) Now slope of PQ = = = − cot ( α / 2 ) cos θ – cos(α – θ) –2sin(α / 2) – sin(θ – α / 2) ⇒ PQ is perpendicular to the line with slope tan(α/2). Hence, Q is the reflection of P in the line through the origin α with slope tan  . 2 MASTERJEE CONCEPTS RELATION BETWEEN THE COEFFICIENT Conditions for two lines to be coincident, parallel, perpendicular and intersecting: Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are a1 b1 c1 Coincident, if = = a2 b2 c2 a1 b1 c1 Parallel, if = ≠ a2 b2 c2 a1 b1 Intersecting, if ≠ a2 b2 Perpendicular, if a1a2 + b1b2 = 0 The three lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 are concurrent, if a1 b1 c1 a2 b2 c2 = 0 a3 b3 c3 Aman Gour (JEE 2012, AIR 230)  M a them a ti cs | 8.27 PROBLEM-SOLVING TACTICS (a) In most of the questions involving figures like triangle or any parallelogram, taking origin as (0,0) helps a lot in arriving at desired solution. One must ensure that conditions given are not violated. (b) One must remember that in an isosceles triangle, centroid, orthocentre, incentre and circumcentre lie on the same line. (c) The centroid, incentre, orthocentre and circumcentre coincide in an equilateral triangle. (d) If area of the triangle is zero, then the three points are collinear. (e) Find the equation of the line passing through two given points, if the third point satisfies the equation of the line, then three points are collinear (f) Whenever origin is shifted to a new point (α, β), then new equation can be obtained by putting x + α in place of x and y + β in placed of y. FORMULAE SHEET (a) Distance Formula: The distance between two points P(x1, y1) and Q(x2, y2) is PQ = (x1 – x2 )2 + (y1 – y 2 )2 = (x1 – x2 )2 + (y1 – y 2 )2 And between two polar co-ordinate A(r1, q1) and B(r2, q2) is AB = r12 + r22 − 2r1r2 cos(θ1 − θ2 ) (b) Section Formula: If P(x1, y1), Q(x2, y2) and the point R(x, y) divide the line PQ internally in the ratio m:n then the co-ordinates of R will be mx2 + nx1 my 2 − ny1  mx + nx1 my 2 + ny1  x= and y = i.e. R  2 ,  m+n m−n ,  m+n m + n   x + x 2 y1 + y 2  And if R is a mid-point of line PQ, then the co-ordinates of R will be  1 ,   2 2  (c) Centroid of Triangle: If A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle ABC and G is  x + x 2 + x3 y1 + y 2 + y 3  Centroid, then co-ordinate of G will be  1 , .  3 3 A   I3 I2 ax1 + bx2 + cx3 ay1 + by 2 + cy 3 (d) Co-ordinates = of Incentre: x = , y B C a+b+c a+b+c (e) Co-ordinates of Ex-centre: As shown in figure, ex-centres of ∆ABC I1 with respect to vertices A, B and C are denoted by I1, I2 and I3, respectively, Figure 8.47  −ax1 + bx2 + cx3 −ay1 + by 2 + cy 3   ax1 – bx2 + cx3 ay1 − by 2 + cy 3  I1 =  ,  ; I2 =  ,  , −a + b + c −a + b + c  a−b + c a−b + c     ax + bx2 − cx3 ay1 + by 2 − cy 3  I3 =  1 ,