Point & Straight Line (Straight Line)

Point & Straight Line STRAIGHT LINE DIFFERENT FORMS OF THE EQUATION A relation between x and y which is satisfied by OF STRAIGHT LINE = ax + by + c = 0 co-ordinates of every point lying on a line is called 1. General form where a , b, c are any real numbers and a, b are not the equation of straight line. Every linear equation in zero both at a time. two variable x and y always represents a straight Particular case ; In ax + by + c = 0 line a=0 by + c = 0 eg. 3x + 4y = 5 which is a line parallel to x axis General form of straight line is given by b=0 ax + c = 0 which is a line parallel to y axis ax + by + c = 0 c=0 ax + by = 0 EQUATION OF THE AXES which is a line passing through origin. (1) Equation of x-axis : y = 0 EQUATION OF STRAIGHT LINE (2) Equation a line parallel to x axis at a distance ‘a’ from it y = a Ex.1 What is the slope of a line whose inclination with the positive direction of x-axis is : (3) Equation of y-axis : x = 0 (i) 0º (ii) 90º (iii) 120º (iv) 150º (4) Equation of a line parallel to y axis at a distance ‘a’ Sol. (i) Here  = 0º from it x = a Slope = tan  = tan 0º = 0. SLOPE OF A LINE (ii) Here  = 90º  The slope of line is not defined. The slope of a line is equal to the tangent of the angle which it makes with the positive side of x-axis (iii) Here = 120º and it is generally denoted by m.  Slope = tan  = tan 120º = tan (180º – 60º) = – tan 60º = – 3 . Thus if a line makes an angle  with x-axis then its (iv) Here  = 150º  Slope = tan  = tan 150º = tan (180º – 30º) = – tan 30º = – The slope of a line joining two points (x1, y1) and (x2 , y2) is given by m = Note : (i) (ii) Slope of x axis or a line parallel to x axis is tan 0° = 0 (iii) Slope of y axis or a line parallel to y axis is tan 90° = 1 3 Ex.2 Find the slope of the line passing through the points : (i) (1, 6) and (– 4, 2) (ii) (5, 9) and (2, 9) Sol. (i) Let A = (1, 6) and B = (– 4, 2)  (ii) Let  4 4 26 = = 5 5  4 1   y  y1  Using slope  2  x 2  x1   Slope of AB = A = (5, 9), B = (2, 9) Slope of AB = 9 9 25 = 0 3 =0 9 2. 4. Point -Slope form : y  y1 = m (x  x1) is the equation of a straight line whose slope is m & which passes through the point (x1, y1). Ex.3 Find the equation of a line passing through (2, –3) and inclined at an angle of 135º with the positive direction of x-axis. Sol. Here, m = slope of the line = tan 135º = tan (90º + 45º) = – cot 45º = –1, x1 = 2, y1 = –3 So, the equation of the line is y – y1 = m (x – x1) i.e.y – (–3) = –1 (x – 2) or y + 3 = – x + 2 or x + y + 1 = 0 3. Point & Straight Line Two point form : The equation of a line passing through two given points ( x1 , y1) and (x2 – y2) is y – y1 = Ex.6 Equation of a line passing through ( 3 , – 4) and ( 4, 3) is [1] y = 7x + 25 [2] y = 7x – 25 [3] y = –7x – 25 [4] y = –7x + 25 Sol. y + 4 = (x–3) or y = 7x – 25 Ans.[2] Ex.7 Find the equation of the line joining the points (– 1, 3) and (4, – 2) Sol. Here the two points are (x1, y1) = (–1, 3) and (x2, y2) = (4, –2). Slope form :So, the equation of the line in two-point form is Where ‘m’ is the slope of the line and ‘c’ is the length 3  ( 2) of the intercept made by it on y axis y–3= (x + 1)  y – 3 = – x – 1  1 4 for general form ax + by + c = 0  x+y–2=0 slope = 5. Determinant form : Equation of line passing through x x (x1, y1) and (x2, y2) is 1 x2 y 1 y1 1  0 y2 1 Ex.8 Find the equation of line passing through (2, 4) & (– 1, 3). x 2 y 1 4 1 x – 3y + 10 = 0 Ex.4 Find the equation of a line with slope –1 and Sol.  1 3 1 = 0  cutting off an intercept of 4 units on negative direction of y-axis. 6. Intercept form : Sol. Here m = –1 and c = – 4. So, the equation of the It is the equation of straight line which cuts off line is y = mx + c intercepts a and b on the axes of x and y respectively. i.e. y = – x – 4 or x + y + 4 = 0 (a) length of the intercept of the line between coordinate axes AB = Ex.5 Equation of a line which is passing through ab (3 , –4) and making an angle of 45° with x axis is (b) Area of  OAB = 2 [1] x – y – 7 = 0 [2] x + y + 7 = 0 (c) Coordinates of the points P which divides AB in the [3] x – y + 7 = 0 [4] x + y – 7 = 0 ratio m1 : m2 are Sol. y – (–4) = tan 45°( x–3) y+4=x–3 x–y–7=0 Ans.[1] 10 Point & Straight Line Ex.12 Equation of a line on which length of pependicular from origin is 4 and inclination of this perpendicular Ex.9 Equation of a line which makes intercepts 3 and is 60° with the positive direction of x axis is 4 on x axis and y axis respectively is [1] x  y 3  8 [2] x  y 3  8 [1] 4x + 3y = 12 [2] 3x + 4y = 12 [3] y  x 3  8 [4] y  x 3  8 [3] 6x + y = 12 [4] 4x – 3y = 12 Sol. x cos60° + y sin60° = 4 Sol. 4x + 3y = 12 Ans.[1] Ans.[1] Ex.10 Find the equation of the line which passes through the point (3, 4) and the sum of its intercepts 8. Parametric form on the axes is 14. x y Sol. Let the equation of the line be a + b = 1 ....(i) This passes through (3, 4), therefore It is given that a + b = 14  3 a + 4 b = 1 ..(ii) b = 14 – a. 3 4 Putting b = 14 – a in (ii), we get a + 14  a = 1  a2 – 13a + 42 = 0  (a – 7) (a – 6) = 0  a = 7, 6 For a = 7, b = 14 – 7 = 7 and for a = 6, b = 14 – 6 = 8. Putting the values of a and b in (i), we get the equations of the lines x y x y + = 1 and 6 + 8 = 1 7 7 or x + y = 7 and 4x + 3y = 24 7. It is the equation of a straight line passes through a given point A (x1 , y1) and makes an angle  with x axis. The coordinates (x , y) of any point P on this line are (x1 + r cos, y1 + r sin) . The distance of this point P from the given point A is Ex.13 Equation of a line which passes through point A(2,3) and makes an angle of 45° with x axis. If this line meet the line x + y + 1 = 0 at point P then distance AP is [1] 2 3 [2] 3 2 [3] 5 2 [4] 2 5 Perpendicular/Normal form : xcos + ysin = p (where p > 0, 0  < 2 ) is Sol. Here x1 = 2 , y1 = 3 and  = 45° the equation of the straight line where the length of hence cos 45 = sin 45 = r the perpendicular from the origin O on the line is p and this perpendicular makes an angle  with positive from first two parts x–2=y–3 xaxis. x–y+1=0 Ex.11 Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the origin to the line makes an angle of 30º with the positive direction of the x-axis. Sol. Here p = 3,  = 30º  Equation of the line in the normal form is x cos 30º + y sin 30º = 3 or x Co-ordinate of point P on this line is if this point is on line x + y + 1 = 0 then +1=0 r=–3 2 ; |r|=3 2 Ans.[2] y 3 + = 3 or 2 2 3 x+y=6 11 Point & Straight Line Ex.14 Find the equation of the line through the point A(2, 3) and making an angle of 45º with the x-axis. Also determine the length of intercept on it between A and the line x + y + 1 = 0 by , y3 = sin 45 º or x2 1 = y 3 1 or 2 2 x–y+1=0 Suppose this line meets the line x + y + 1 = 0 at P such that AP = r. Then the co-ordinates of P are given by x2 y3 = =r cos 45 º sin 45 º  x = 2 + r cos 45º, y = 3 + r sin 45º  r x=2+ ,y=3+ 2 r  2 r   2 ,3 Since P lies on x + y + 1 = 0, so 2 + r 2 +3+ r 2 +1=0 intercept c = 2. 5 y-intercept = 5 3 tan = If the given equation of straight line are a1x + b1y + c1 = 0 and a2x + b2 y + c2 =0 , then the angle between the lines ‘’ is given by a2 b1 – a1 b2 = 0 or (ii) If the lines are perpendicular, then a1a2 + b1b2 = 0 , (iii) Coincident :(iv) In particulars, angle between lines a1 x + b1 y + c1 Intercept form is = 1 , here x intercept is = 3. 2 3 (i) If the lines are parallel, then b y intercept is = = tan = General form of equation ax + by +c = 0 then its 1. Slope intercept form is , here slope m = a b ANGLE BETWEEN TWO LINES 1. The angle  between two lines whose slopes are m1 and m2 is given by 2. 2 REDUCTION OF GENERAL FORM OF EQUATION INTO STANDARD FORM y= slope = – (i) If the lines are parallel, then m1 = m2 (ii) If the lines are perpendicular, then m1 m2 = –1 r = – 6  r = –3 2  length AP = | r | = 3 2 Thus, the length of the intercept = 3 2 .   c 2  Ex.15 Find slope, x-intercept & y-intercept of the line 2x – 3y + 5 = 0. Sol. Here, a = 2, b = – 3, c = 5 x-intercept = – a = – 2 r Thus, the co-ordinates of P are  2  a here cos = Sol. The equation of a line through A and making an angle of 45º with the x-axis is x2 cos 45 º like = 0 and b2y + c2 = 0 is tan–1 , a1 b1 and angle between lines a1x + b1y + c1 = 0 and a2x + c2 = 0 is tan–1 b1 a1 Normal form is :- To change the general form of a (v) Angle between lines line into normal form, first take c to right hand side x cos  + y sin  = p1 and x cos  + y sin  = p2 and make it positive, then divide the whole equation = 12 Point & Straight Line 3m  1 3 m Ex.16 The acute angle between two lines is /4 and slope of one of them is 1/2. Find the slope of the other line. Sol. If  be the acute angle between the lines with slopes  The slope of the required line is not defined. Thus, the required line is a vertical line. This line is to pass through the origin. m1  m2  4  1 2 and m1 = [1] tan 1 Sol. Ex.17 Find the equation of the straight line which passes through the origin and making angle 60º with the line Slope of (1) = – Sol. 3 1 3 3m  1 3 m  3m  1 m=  3 m = 3  tan 2 [2] tan–1 1 [4] none of these = tan–1 2 (use form (iv) ) Ans.[3] y= is here m1 = 1 m2 = [1] 75 [2] 45 3m  1 3 m 3 m+1=3– 1 =± 3 3m 3 [3] 15 [4] 30  = tan–1 Sol. tan = tan 60º = 1  m  1/ 3 3 = 1 2 Ex.20 Angle between y = x + 6 and . m   1/ 3  = tan–1 (1/2) (use form (iv)) [3] tan–1 2 Let slope of the required line be m. Also between these lines is given to be 60º.  2 [1] tan 1 Sol. Given line is x + 3 y + 3 3 = 0.  tan [4] none of these Ex.19 The angle between the lines 2x – y + 1 = 0 and 2y + 1 = 0 is given by x + 3 y + 3 3 = 0.  [2] tan–1 1 Ans.[1] = + 1 or – 1  1  y =    x – 3 1 2 [3] tan–1 2 1  2m2  1= 2  m2  The equation of the required line is x = 0 Ex.18 The angle between the lines 2x – y + 1 = 0 and x + 5 = 0 is given by 1  m2 2   tan = 1 1  m2 4 2 1  2m 2  2m 2 m is not defined  m1 and m2, then tan  = 1  m m 1 2 Let  = =– 3  3m+1 =–3+ 3 m Ans.[3] Ex.21 If A(1,2), B(–1, 3) and C (3, –5) be vertices of a triangle then find [1] tan 1 1 2 [3] tan–1 2 3 [2] tan–1 1 [4] tan 1 3 4 Using y = mx + c, the equation of the required line is Sol. Slope of AB = m1 = –1/2 Slope of BC = m2 = –2 1 y= x+0 3 = tan–1 (3/4) Ans.[4] i.e.x – 3 y = 0. ( This passes through origin, so c = 0) 13 Point & Straight Line Ex.22 If x + 4y – 5 = 0 and 4x + ky + 7 = 0 are two perpendicular lines then k is (1) 3 (2) 4 (3) –1 (4) –4 Sol. m1 = m2 = Two lines are perpendicular if m1 m2 = –1 k=–1 Ex.24 Find the equation of the straight line that has y-intercept 4 and is parallel to the straight line 2x – 3y = 7. Sol. Given line is 2x – 3y = 7 (1) Ans.[3] 3y = 2x – 7   2 7 y= 3 x – 3 Slope of (1) is 2/3 The required line is parallel to (1), so its slope is also 2/3, y-intercept of required line = 4  By using y = mx + c form, the equation of the required line is  Ex.23 If 7x + 3y + 9 = 0 and y = kx + 7 are two parallel lines then k is (1) 3/7 (2) –7/3 (3) 3 (4) 7 Sol. m1 = m2 = k 2 Two lines are parallel if m1 = m2 ; k = –7/3 Ans.[2] y = 3 x + 4 or 2x – 3y + 12 = 0 Ex.25 Two sides of a square lie on the lines x + y = 1 and x + y + 2 = 0. What is its area ? Parallel Lines : (i) When two straight lines are parallel their slopes are Sol. Clearly the length of the side of the square is equal to the distance between the parallel lines equal. Thus any line parallel to x+y–1=0 ........(i) and y = mx + c is of the type y = mx + d, where ‘d’ is a parameter. x+y+2=0 ........(ii) (ii) Two lines ax + by + c = 0 and ax + by + c = 0 Putting x = 0 in (i), we get y = 1. So (0, 1) is a point on line (i). are parallel if =  .    Now, Distance between the parallel lines Thus any line parallel to ax + by + c = 0 is of the type ax + by + k = 0, where k is a parameter. (iii) The distance between two parallel lines with equations ax + by + c1 = 0 & ax + by + c2 = 0 is = c 1 c 2 a 2 b 2 = length of the  from (0, 1) to x + y + 2 = 0 | 0  1 2 | = . 2 2 1 1 3 = 2 3 2 Thus, the length of the side of the square is 3 2 and 2  3  9 hence its area =   = 2  2 Note that coefficients of x & y in both the equations must be same. p1p 2 (iv) The area of the parallelogram = sin , where p1 & p2 are distances between two pairs of opposite sides &  is the angle between any two adjacent sides. Note that area of the parallelogram bounded by the lines y = m1x + c1, y = m1x + c2 and y = m2x + d1, y = m2x + d2 is given Ex.26 Find the area of the parallelogram whose sides are x + 2y + 3 = 0, 3x + 4y – 5 = 0, 2x + 4y + 5 = 0 and 3x + 4y – 10 = 0 Sol. ( c c )( d  d ) by 1 2 1 2 m 1 m 2 14 3 5 Here, c1 = – , c2 = – , 2 4 10 5 1 3 d1 = , d2 = , m 1 = – , m 2 = – 4 4 2 4  3 5   10 5         2 4 4 4 5  Area = = sq. units  1 3 4     2 4 Ex.27 The distance between the lines 3x + 4y – 5 = 0 and 3x + 4y + 7 = 0 [1] 12 [2] 2 [3] 12/5 [4] 5/12 Sol. = Point & Straight Line Sol. The equation of a line perpendicular to 3x + 2y + 5 = 0 is 2x – 3y +  = 0 ...........(i) This passes through the point (3, 4)  3×2–3×4+=0=6 Putting = 6 in (i), we get 2x – 3y + 6 = 0, which is the required equation. Aliter The slope of the given line is –3/2. Since the required line is perpendicular to the given line. So, the slope of the required line is 2/3. As it passes through 2 (3, 4). So, its equation is y – 4 = 3 (x – 3) or 2x – 3y + 6 = 0 Ex.28 Equation of a line which passes through (4,6) and parallel to 3x – 7y + 2 = 0 is (1) 3x – 7y + 30 = 0 (3) 3x – 7y – 30 = 0 (2) 3x + 7y + 30 = 0 (4) 3x + 7y – 30 = 0 Sol. Let the equation is 3x – 7y + k = 0 this line passes through ( 4, 6) Hence 3(4) – 7 (6) + k = 0 k = 30 the requiired equation is 3x – 7y + 30 = 0 Ans.[1] POSITION OF A POINT RELATIVE TO A LINE (i) The point (x1 , y1) lies on the line ax + by + c =0 if, ax1 + by1 + c = 0 (ii) If P (x1, y1) and Q (x2 , y2) do not lie on the line ax + by +c = 0 then they are on the same side of the line, if ax1 + by1 + c and ax2 + by2 + c are of the same sign and they lie on the opposite side of line if ax1 + by1 + c and ax2 + by2 + c are of the opposite sign. Perpendicular Lines: (iii) (x1 , y1) is on origin or non origin sides of the line ax (i) When two lines of slopes m1 & m2 are at right angles, + by + c = 0 if ax1 + by1 + c and c are of the same the product of their slopes is 1, or opposite signs. i.e. m1 m2 =  Thus any line perpendicular to y = mx + c is of the form y=  x + d, where ‘d’ is any parameter.. (ii) Two lines ax + by + c = 0 and ax + by + c = 0 are perpendicular if aa + bb = 0. Thus any line perpendicular to ax + by + c = 0 is of the form bx  ay + k = 0, where ‘k’ is any parameter. Ex.30 Points (3,4) and (–9,6) lie the line 7x + 5y – 9 = 0 [1] on same side [2] on opposite side [3] on the line [4] nothing can be predicted Sol. Putting both the points in the given equation Ex.29 Find the equation of the straight line that passes through the point (3, 4) and perpendicular to the line 3x + 2y + 5 = 0 7 × 3 + 5 × 4 – 9 = 32 and 7 × (–9) + 5(6) – 9 = – 42 as both are of opposite sign so they lie opposite side of the given line Ans.[2] 15 Point & Straight Line Ex.31 Show that (1, 4) and (0, –3) lie on the opposite sides of the line x + 3y + 7 = 0. Sol. At (1, 4), the value of x + 3y + 7 = 1 + 3(4) + 7 = 20 > 0. At (0, – 3), the value of x + 3y + 7 = 0 + 3(–3) + 7 = –2 < 0 The points (1, 4) and (0, – 3) are on the opposite sides of the given line.  EQUATION OF A LINE MAKING A GIVEN ANGLE WITH ANOTHER LINE The equation of a line passing through (x1, y1) and making an angle  with the line y = mx + c is given by y  y1  m  tan  ( x  x1 ) 1  m tan  The ratio in which a given line divides the line segment joining two points : Let the given line ax + by + c = 0 divide the line segment joining A (x 1, y1) & B(x2, y2) in the ratio Ex.33 One vertex of an equilateral triangle is (2,3) and the equation of line opposite to the vertex is x + y = 2, then equation of remaining two sides are same side of the given line then m/n is negative but if (1) y – 3 = 2 ( x – 2) A & B are on opposite sides of the given line, then (2) y – 3 = ( x –2) m/n is positive m : n, then m ax1by1c  . If A & B are on the n ax 2 by 2 c (3) y – 3 = ( x – 2) (4) None of these Ex.32 Find the ratio in which the line joining the points Sol. Since the two sides make an angle of 60° each with side x + y = 2. Therefore equation of these sides will A (1, 2) and B(– 3, 4) is divided by the line x + y – be 5 = 0. Sol. Let the line x + y = 5 divides AB in the ratio k : 1 at P    3k  1 4k  2  ,  k 1   k 1 co-ordinate of P are  Since P lies on x + y – 5 = 0    3k  1 4k  2 1 + –5=0  k=– k 1 k 1 2 Required ratio is 1 : 2 extrenally. Aliter Let the ratio is m : n (1 1  1 2  5) m 1  = – 1 ( 3)  1 4  5 = – n 2  ratio is 1 : 2 externally. 1  tan 60 y – 3 = 1  ( 1) tan 60 ( x  2) = y–3 =(2 ) ( x – 2) Ans.[3] CONCURRENCY OF THREE LINES Three lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 a1 b1 c1 are concurrent iff a 2 b 2 c 2  0 a3 b3 c3 Note : If P = 0, Q = 0, R = 0 are equations of three lines, then P + Q + R = 0  the lines are concurrent. But its converse may not be true i.e., if the three lines are concurrent then it is not necessary that P + Q + R = 0. 16 Point & Straight Line 4 t  3( 4  t )  10  Ex.34 If the lines x + y = 1, 2x – y = 0 and x + 2y +  = 0 are concurrent then find . [1] –5/3 [2] 5/3 [3] 3/5 [4] –3/5 Sol. Using condition of concurrency, we have 1 1 1 2 1 0  0 1 2  1(–) –1(2) –(4 + 1) = 0  –3 – 5 = 0   = –5/3. The length of perpendicular from P(x1, y1) on ax + by + c = 0 is |t + 2| = 5 t+2=±5  t = –7 or t = 3 Hence, required points are (–7, 11) and (3, 1) Self practice problem : Find the length of the altitudes from the vertices of the triangle with vertices :(–1, 1), (5, 2) and (3, –1). 16 13 , 8 5 , 16 37 Ans.[1] Length of perpendicular from a point on a line :  =1   Answer :   42  3 2 .  Reflection of a point about a line : (i) Foot of the perpendicular from a point (x1, y1) on the line ax + by + c = 0 is  ax  by1  c  x  x1 y  y1      1 2  a b a  b2   (ii) The image of a point (x1, y1) about the line ax + by + c = 0 is  ax  by1  c  x  x1 y  y1 .   2  1 2  a b a  b2   Ex.35 Find the distance between the line 12x – 5y + 9 = 0 and the point (2, 1) Sol. The required distance = = | 24  5  9 | 13 = Ex.37 Find the foot of perpendicular of the line drawn from P (– 3, 5) on the line x – y + 2 = 0. Sol. Slope of PM = – 1 12  2  5  1  9 12 2  ( 5 )2 28 13 Ex.36 Find all points on x + y = 4 that lie at a unit distance from the line 4x + 3y – 10 = 0. Sol. Note that the co-ordinates of an arbitrary point on x + y = 4 can be obtained by putting x = t (or y = t) and then obtaining y (or x) from the equation of the line, where t is a parameter. Putting x = t in the equation x + y = 4 of the given line, we obtain y = 4 – t. So, co-ordinates of an arbitrary point on the given line are P(t, 4 – t). Let P(t, 4 – t) be the required point. Then, distance of P from the line 4x + 3y – 10 = 0 is unity i.e. Equation of PM is x+y–2=0 .........(i) solving equation (i) with x – y + 2 = 0, we get co-ordinates of M (0, 2) Aliter  Here, (1 ( 3)  ( 1)  5  2) x3 y5 = =– (1)2  (1) 2 1 1 x3 y5 = =3 1 1  x+3=3   y–5=–3  M is (0, 2)  x=0 y= 2 and 17 Point & Straight Line Ex.38 Find the image of the point P(–1, 2) in the line mirror 2x – 3y + 4 = 0. Self practice problems : 1. Find the foot of perpendicular of the line drawn from (– 2, – 3) on the line 3x – 2y – 1 = 0. 2. Find the image of the point (1, 2) in y-axis. Sol. Let image of P is Q. Answers (27)     23  41  ,   13   13 (28) (– 1, 2) PM = MQ & PQ  AB BISECTORS OF THE ANGLES BETWEEN Let Q is (h, k) TWO LINES M is  h 1 k  2  ,   2   2 Bisectors of the two angles between two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are given by It lies on 2x – 3y + 4 = 0.  h  1 k  2 2  2  – 3  2  + 4 = 0.     or 2h – 3k = 0 ...........(i)  k2 h1 slope of PQ = PQ  AB  2 k2 × 3 = – 1. h1  3h + 2k – 1 = 0. ........(ii) soving (i) & (ii), we get h =   3 3 2 ,k= 13 13 2  Image of P(– 1, 2) is Q  13 , 13    Aliter First write the equation of the lines so that the constant terms are positive. Then (a) If a1a2 + b1b2 > 0 then on taking positive sign in the above bisectors equation we shall get the obtuse angle bisector and on taking negative sign we shall get the acute angle bisector. The image of P (– 1, 2) about the line 2x – 3y + 4 = 0 (b) If a1a2 + b1b2 < 0 then positive sign give the acute angle bisector and negative sign gives the obtuse [ 2(1)  3(2)  4] y2 x 1 is = 3 =– 2 2 2 2  (3) 2 angle bisector. x 1 = 2  y2 3 = 8 13 13x + 13 = 16 13y – 26 = – 24   3 3 13  x=  y = 13 & (c) On taking positive sign we shall get equation of the bisector of the angle which contains the origin and negative sign gives the equation of the bisector which does not contain origin. 2 2  image is  13 , 13    18 Point & Straight Line Sol. (i) The equations of the given straight lines are Ex.39 Find the equation of the bisector of the acute angle between the lines 3x – 4y + 7 = 0 and 12x + 5y – 2 =0 [1] 11x – 3y + 9 = 0 [3] 11x – 3y – 9 = 0 [2] 9x – 11y + 9 = 0 [4] 11x + 3y – 9 = 0 Here a1a2 + b1b2 = 3(–12) –4)(–5)< 0, so positive sign gives the acute angle bisector which is 99x – 27y + 81 = 0 Ans.[1] 11x – 3y + 9 = 0 Note : This is also the bisector of the angle in which origin lies ( since c1 , c2 are positive and it has been obtained by taking positive sign) Ex.40 Find the equations of the bisectors of the angle between the straight lines 3x – 4y + 7 = 0 and 12 x – 5y – 8 = 0. Sol. The equations of the bisectors of the angles between 3x – 4y + 7 = 0 and 12 x – 5y – 8 = 0 are 3x  4y  7 3 2  ( 4 ) 2 or =± 3x  4y  7 5 12 x  5y  8 =± 12 2  ( 5) 2 12 x  5y  8 13 or 39x – 52y + 91 = ± (60 x – 25 y – 40) Taking the positive sign, we get 21 x + 27 y – 131 = 0 as one bisector Taking the negative sign, we get 99 x – 77 y + 51 = 0 as the other bisector. Ex.41 For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, find the equation of the (i) bisector of the obtuse angle between them; (ii) bisector of the acute angle between them; ........(1) 5x + 12y + 9 = 0 ........(2) The equation of the bisectors of the angles between lines (1) and (2) are 4x  3y  6 2 Sol. The given equation can be written as 3x – 4y + 7 = 0 and –12x – 5y + 2 = 0 4x + 3y – 6 = 0 4 3 2 4x  3y  6 5 5 x  12y  9 =± =± 5 2  12 2 5 x  12 y  9 13 4x  3y  6 5 Taking the positive sign, we have = or 5 x  12 y  9 13 or 52x + 39y – 78 = 25x + 60y + 45 or 27x – 21y – 123 = 0 or 9x – 7y – 41 = 0 Taking the negative sign, we have =– 4x  3y  6 5 5 x  12 y  9 13 or 52x + 39y – 78 = –25x – 60y – 45 or 77x + 99y – 33 = 0 or7x + 9y – 3 = 0 Hence the equation of the bisectors are 9x – 7y – 41 = 0 ........(3) and 7x + 9y – 3 = 0 ........(4) 4 Now slope of line (1) = – 3 and slope of the bisector (3) = 9 . 7 If  be the acute angle between the line (1) and the bisector (3), then 9 4  27  28 7 3 tan  = = 21  36 9 4 1    7 3   > 45º 55 11 =  15 = 3 > 1 Hence 9x – 7y – 41 = 0 is the bisector of the obtuse angle between the given lines (1) and (2) (ii) Since 9x – 7y – 41 is the bisector of the obtuse angle between the given lines, therefore the other bisector 7x + 9y – 3 = 0 will be the bisector of the acute angle between the given lines. 19 Point & Straight Line 2nd Method :        = according as Writing the equation of the lines so that constants     become positive we have a  + b  + c and a  + b  + c having same sign – 4x – 3y + 6 = 0 .......(1) or otherwise. and 5x + 12y + 9 = 0 .......(2) Here a1 = – 4, a2 = 5, b1 = –3, b2 = 12 Now a1a2 + b1b2 = – 20 – 36 = –56 < 0 Ex.42 For the straight lines 4x + 3y – 6 = 0 and 5x +  origin does not lie in the obtuse angle between 12y + 9 = 0, find the equation of the bisector of the lines (1) and (2) and hence equation of the bisector angle which contains the origin. of the obtuse angle between lines (1) and (2) will be Sol. For point O(0, 0), 4x + 3y – 6 = –6 < 0 and 5x + 4 x  3y  6 5 x  12y  9 12y + 9 = 9 > 0 =– (  4 ) 2  (  3) 2 5 2  12 2 Hence for point O(0, 0) 4x + 3y – 6 and 5x + 12y + or 13(–4x – 3y + 6) = –5(5x + 12y + 9) 9 are of opposite signs. or 27x – 21y – 123 = 0 or 9x – 7y – 41 = 0 Hence equation of the bisector of the angle between the given lines containing the origin will be and the equation of the bisector of the acute angle 4x  3y  6 will be (origin lies in the acute angle) 5 x  12y  9 2 2 = – 4 x  3y  6 ( 4 )  (3 ) 5 x  12 y  9 5 2  12 2 = 4x  3y  6 5 x  12 y  9 (  4 ) 2  (  3) 2 5 2  12 2 or =– 5 13 or 77x + 99y – 33 = 0 or 7x + 9y – 3 = 0 or 52x + 39y – 78 = –25x – 60y – 45. To discriminate between the bisector of the angle or 77x + 99y – 33 = 0 or 7x + 9y – 3 = 0 containing a point : LINES PASSING THROUGH THE POINT OF To discriminate between the bisector of the angle INTERSECTION OF TWO LINES containing the origin & that of the angle not containing If equation of two lines L1 = a1x+ b1y + c1 = 0 and the origin. Rewrite the equations, ax + by + c = 0 & L2 = a2x + b2y + c2 = 0 then the equation of the lines ax + by + c= 0 such that the constant terms c, c passing through the point of intersection of these lines is L1 + L2 = 0 or (a1x + b1y + c = 0) +  (a2x + b2y are positive. Then ; + c2 = 0) = 0, value of  is obtained with the help of  the additional infomation given in the problem. =+ gives the equation of the    Ex.43 Equation of a line passing through the point of intersection of x + y – 3 = 0 and 2x – y + 1= 0 and a point (2, –3) is & =  gives the     [1] 4x + y – 5 = 0 [2] 4x – y – 5 = 0 [3] 4x + 5y + 5 = 0 [4] 4x + 2y – 5 = 0 equation of the bisector of the angle not containing Sol. (x + y – 3) +  (2x – y + 1) = 0 the origin. In general equation of the bisector which as this line also passes through ( 2, –3) contains the point (, ) is, Hence (2 – 3 – 3) +  ( 2 × 2 – (–3) +1) = 0        – 4 + 8 = 0 ;  = 1/2 = or equation is (x + y – 3) + 1/2 (2x – y +1 ) = 0     4x + y – 5 = 0 Ans.[1] 20 bisector of the angle containing the origin Point & Straight Line Ex.44 Find the equation of the straight line which passes A Pair of straight lines through origin: through the point (2, –3) and the point of intersection (i) A homogeneous equation of degree two, of the lines x + y + 4 = 0 and 3x – y – 8 = 0. "ax² + 2hxy+ by² = 0" always represents a pair of Sol. Any line through the intersection of the lines x + y + straight lines passing through the origin if : 4 = 0 and 3x – y – 8 =0 has the equation (a) h² > ab  lines are real & distinct . (x + y + 4) +  (3x – y – 8) = 0 .........(i) (b) h² = ab  lines are coincident . This will pass through (2, –3) if (2 – 3 + 4) +  (6 + 3 – 8) = 0 or 3 +  = 0   = – 3. Putting the value of  in (i), the required line is (x + y + 4) + (–3) (3x – y – 8) = 0 or – 8x + 4y + 28 = 0 or 2x – y – 7 = 0 Aliter Solving the equations x + y + 4 = 0 and 3x – y – 8 = 0 by cross-multiplication, we get x = 1, y = –5 So the two lines intersect at the point (1, –5). Hence the required line passes through (2, –3) and –5  3 (1, –5) and so its equation is y + 3 = (x – 2) 1 2 or 2x – y – 7 = 0 Ex.45 Obtain the equations of the lines passing through the intersection of lines 4x – 3y – 1 = 0 and 2x – 5y + 3 = 0 and equally inclined to the axes. Sol. The equation of any line through the intersection of the given lines is (4x – 3y – 1) +  (2x – 5y + 3) = 0 or x (2  + 4) – y (5 + 3) + 3 – 1 = 0 (c) h² < ab  lines are imaginary with real point of intersection i.e. (0, 0) This equation is obtained by multiplying the two equations of lines (m1x – y) (m2x – y) = 0  m1m2x2 – (m1 + m2) xy + y2 = 0 (ii) If y = m1x & y = m2x be the two equations represented by ax² + 2hxy + by² = 0, then; m 1 + m2 =  & m1 m2 = . (iii) If  is the acute angle between the pair of straight lines represented by, 2 h 2  ab ax² + 2hxy + by² = 0, then tan = ab . .......(i) (iv) The condition that these lines are : (a) at right angles to each other is a + b = 0. Let m be the slope of this line. Then m = i.e. coefficient of x² + coefficient of y² = 0. As the line is equally inclined with the axes, therefore (b) coincident is h² = ab . m = tan 45º or m = tan 135º (c) equally inclined to the axis of x is h = 0.i.e. coeff. 2  4 of xy = 0 .  m = ±1, =±1 2  4 5  3 5  3 1 = –1 or 3 , putting the values of  in (i), we get 2x + 2y – 4 = 0 and 14x – 14y = 0   i.e.x + y – 2 = 0 and x = y as the equations of the required lines. Note that a homogeneous equation of degree n represents n straight lines passing through origin. (v) The equation to the pair of straight lines bisecting the angles between the straight lines ax² + 2hxy + by² = 0 is   = . 21 Ex.46 Show that the equation 6x2 – 5xy + y2 = 0 represents a pair of distinct straight lines, each passing through the origin. Find the separate equations of these lines. Sol. The given equation is a homogeneous equation of second degree. So, it represents a pair of straight lines passing through the origin. Comparing the given equation with ax2 + 2hxy + by2 = 0, we obtain a = 6, b = 1 and 2h = – 5. 25 1  h – ab = – 6 = > 0  h2 > ab 4 4 2 Point & Straight Line Ex.48 Find the angle between the pair of straight lines 4x2 + 24xy + 11y2 = 0 Sol. Given equation is 4x2 + 24xy + 11y2 = 0 Here a = coeff. of x2 = 4, b = coeff. of y2 = 11 and 2h = coeff. of xy = 24  h = 12 Now tan  = 2 h2  ab ab = 2 144  44 4  11 = 4 3 Where  is the acute angle between the lines.  acute angle between the lines is tan–1  4  and obtuse angle between them is  – tan–1 3 4   3 Hence, the given equation represents a pair of distinct Ex.49 Find the equation of the bisectors of the angle lines passing through the origin. between the lines represented by Now, 6x2 – 5xy + y2 = 0 3x2 – 5xy + 4y2 = 0 2 y y    –5   +6=0 Sol. Given equation is 3x2 – 5xy + 4y2 = 0 .......(1) x x comparing it with the equation ax2 + 2hxy + by2 = 0 2 y y y .......(2)    –3   –2   +6=0 x x x we have a = 3, 2h = –5; and b = 4 y  y   3  2     Now the equation of the bisectors of the angle  =0 x x  y  x    y – 3 = 0 or – 2 = 0  y – 3x = 0 or y – 2x = 0 x between the pair of lines (1) is x2  y2 xy = ab h So the given equation represents the straight lines 2xy x2  y 2 xy x2  y 2 or = ; or = y – 3x = 0 and y – 2x = 0. 5 5 34  1  2 Ex.47 Find the equations to the pair of lines through the origin which are perpendicular to the lines or 5x2 – 2xy – 5y2 = 0 represented by 2x2 – 7xy + 3y2 = 0. Ex.50 Lines represented by 2x2 – 7xy + 3y2 = 0 are Sol. We have 2x2 – 7xy + 3y2 = 0. [1] x + 2y = 0, x – 3y = 0  2x2 – 6xy – xy + 3y2 = 0 [2] x – 2y = 0, x + 3y = 0  2x(x – 3y) – y (x – 3y) = 0 [3] 2x – y = 0, 3x – y = 0  (x – 3y) (2x – y) = 0 [4] 2x – y = 0, x – 3y = 0  x – 3y = 0 or 2x – y = 0 2 2 Thus the given equation represents the lines x – 3y Sol. 2x – 7xy + 3y = 0 = (2x – y) (x – 3y) = 0 and 2x – y = 0. The equations of the lines passing 2x – y = 0, x – 3y = 0 Ans.[4] through the origin and perpendicular to the given lines Ex.51 Angle between the lines 2x2 – 7xy + 3y2 = 0 is 1 are y – 0 = –3 (x – 0) and y – 0 = – (x – 0) 2 [1] 45 [2] 30 [3] 60 [4] 90 [  (Slope of x – 3 y = 0) is 1/3 and (Slope of 2x – y = 0) is 2] Sol. tan tan = 1  = 45°  y + 3x = 0 and 2y + x = 0 Ans.[1] 22 Point & Straight Line General equation of second degree representing a pair of Straight lines : EQUATION OF LINES JOINING THE INTERSECTION POINTS OF A LINE & (i) ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a A CURVE TO THE ORIGIN pair of straight lines if : Let lx + my + n = 0 ....... (1) and the second degree a h g curve ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ....(2) abc + 2fgh  af²  bg²  ch² = 0, i.e. if h b f = 0. then their joint equation is g f c Such an equation is obtained again by multiplying the two equation of lines ax 2 + 2hxy + by 2 + 2gx  +c (a1x + b1y + c1) (a2x + b2y + c2) = 0   + 2fy =0 (ii) The angle  between the two lines representing by a general equation is the same as that between the two lines represented by its homogeneous part only. i.e. making the equation (2) homogenous using equation (1) Ex.52 Prove that the equation 2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0 represents a pair of straight lines. Find the co-ordinates of their point of intersection. Sol. Given equation is 2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0 Ex.53 Equation of pair of straight lines which are formed by joining the origin and the points of intersection of a circle x2 + y2 = a2 and a line y = mx + c is Writing the equation (1) as a quadratic equation in x we have 2x2 + (5y + 6) x + 3y2 + 7y + 4 = 0  (5 y  6)  (5 y  6)2  4.2(3 y 2  7 y  4 ) 4  x= =  (5 y  6)  25 y 2  60 y  36  24 y 2  56 y  32 4 =  ( 5 y  6)  y 2  4 y  4 = (5 y  6)  ( y  2) 4 4 5 y  6  y  2 5y  6  y  2  x= , 4 4 or 4x + 4y + 4 = 0 and 4x + 6y + 8 = 0 or x + y + 1 = 0 and 2x + 3y + 4 = 0 [1] x2 (c2 + a2m2 ) + y (c2 + a2) – 2ma2xy = 0 [2] x2 (c2 – a2m2 ) – y (c2 – a2) + 2ma2xy = 0 [3] x2 (c2 – a2m2 ) + y (c2 – a2) – 2ma2xy = 0 [4] x2 (c2 – a2m2 ) + y (c2 + a2) – 2ma2xy = 0 Sol. x2 + y2 = a2 x2 (c2 – a2m2 ) + y (c2 – a2) – 2ma2xy = 0 Ans.[3] Ex.54 For what value of k, the equation kx2 – 10xy + 12y2 + 5x – 16y –3=0 represents a pair of straight lines [1] 2 [2] 3 [3] 8 [4] 4 Hence equation (1) represents a pair of straight lines Sol. Here a = k, b = 12, c = –3 , h = –5, g = 5/2 , f = – 8 whose equation are x+y+1=0 .....(1) and 2x + 3y + 4 = 0 .....(2) Solving these two equations, the required point of intersection is (1, – 2). = k(12) (–3) + 2 (–8) (5/2) (–5) –k(–8) 2 – 12(5/2)2 –(–3) (–5)2 = 0 k=2 Ans.[1] 23 Point & Straight Line Ex.55 The angle and equation of bisectors of angle between the lines represented by the eqation 2x2 – 7xy + 3y2 = 0 are 2 or 7x – 2xy – y = 0 This is the equation of the lines joining the origin to the points of intersection of (1) and (2). [1] 45°, 7x2 – 2xy – 7y2 = 0 2 Comparing equation (3) with the equation ax2 + 2hxy + by2 = 0 2 [2] 60°, 7x – 2xy – 7y = 0 [3] 120°, 7x2 – 2xy – 7y2 = 0 we have a = 7, b = –1 and 2h = –2 [4] 45°, 7x2 + 2xy + 7y2 = 0 given lines are 2x – y = 0 and x – 3y = 0 tan  = since a = 2 , h = – 7/2, b = 3 Therefore angle between lines given by =1   = tan–1 7x2 – 2xy – 7y2 = 0 2 1 7 7 1 = 2 8 2 2 = 6 3 2 2 . 3 Parallel transformation Let origin O 0,0 be shifted to a point a,b by moving the x-axis and y-axis, parallel to themselves. If the co-ordinate of point P with reference to old axis are x1, y1 then co-ordinate of this point with respect to new axis will be Ans.[1] Ex.56 Prove that the angle between the lines joining the origin to the points of intersection of the straight line y = 3x + 2 with the curve x2 + 2xy + 3y2 + 4x + 8y – 11 = 0 is tan–1 = TRANSFORMATION OF AXES 1. xy 2 h 2  ab ab  = 45° Again equations of bisectors are given by = i.e. h = –1 If  be the acute angle between pair of lines (3), then Sol. Since 2x2 – 7xy + 3y2 = (2x – y) (x – 3y) tan 2 2 2 . 3 P (x’ , y’ ) = P (x1–a , y1 –b) Sol. Equation of the given curve is x2 + 2xy + 3y2 + 4x + 8y – 11 = 0 and equation of the given straight line is y – 3x = 2;  y  3x =1 2 Making equation (1) homogeneous equation of the second degree in x any y with the help of (1), we  y  3x    2  have x2 + 2xy + 3y2 + 4x   y  3x    2  + 8y  2  y  3x  – 11  2  = 0   or x2 + 2xy + 3y2 + – 1 (4xy + 8y2 – 12x2 – 24 xy) 2 11 2 (y – 6xy + 9x2) = 0 4 2. Rotational transformation: Let OX and OY be the old axes and OX’ and OY’ be the new axes obtained by rotating the old axes OX and OY through an angle , again, if coordinates of any point P x,y with reference to new axes will be x ', y ' , then or 4x2 + 8xy + 12y2 + 2(8y2 – 12x2 – 20xy) – 11 (y2 – 6xy + 9x2) = 0 or –119x2 + 34xy + 17y2 = 0 or 119x2 – 34xy – 17y2 = 0 24 The above relation between (x , y) and ( x’ , y’) can be easily obtained with the help of following table 3 Point & Straight Line Ex.58 If the new coordinates of a point after the rotation x y of axes in the negative direction by an angle of 3 are 4,2 then coordinates with respect to old axes are x’ cos sin [1] y’ –sin cos [2] Reflection (image) of a point: [3] Let x, y be any point, then its image with respect to [4] (i) x (ii) y Sol. (iii) (iv) y  x 'sin   y 'cos  Ex.57 then new co-ordinates of Sol. is [1] (6, –6) [2] (–6, 6) [3] (4, –4) [4] (6, 6) Hence the coordinates are Ans.[1] Ans.[1] 25