Straight Line Solved Problems

Page # 3 STRAIGHT LINE SOLVED PROBLEMS Ex.1 Sol. Find the equation of a line which is making an angle of 60º with the x axis and an intercept of 5 unit length in negative direction of y axis Let line is y = mx + c  r r   2   ,3  2 2  if this point is on line x + y + 1 = 0 then  r   r   2     3   + 1 = 0 2  2  Here m = tan60º = 3 , c = – 5 equation is y = Ex.2 Sol. Ex.3 Sol. Ex.4 Sol. Ex.5 Sol. Ex.6 Sol.  3x–5 Find the equation of a line which is passing through origin and making an angle of 45º with x-axis Here m = tan45º = 1 so the equation is y = x or x – y = 0 Find the equation of a line which is passing through (3, – 4) and making an angle of 45º with x-axis y – (–4) = tan45º(x – 3)  y+4=x–3 x–y–7=0 Ex.7 Sol. Ex.8 Find the equation of a line passing through (3, –4) and (4, 3) 3 4 y+4= (x–3) or y = 7x – 25 43 Find the equation of a line which makes intercepts 3 and 4 on x-axis and y-axis x y  =1 3 4 hence x2 y3 = =r cos 45º sin 45º y r A(2, 3) P x 45º x O x + y+ 1 = 0 y from first two parts  x – 2 = y – 3 x–y+1=0 Co-ordinate of point P on this line is | r |= 3 2 Find the equation of a line which passes through (1, –2) and makes equal intercept on axes x y Let the equation is  = 1, given a b a = b  Line passes through (1, 2) 1 2   = 1  a = –1 a a x y  equation is =1 1 1 x+y+1=0 Find the slope of a Line which is made by joining the point (1, 0) and (–2, 3) 1 3 0 =– 3  2 1 1 tan = –  = 150º 3 Sol. m= Ex.9 If the intercept made by the line between the axes is bisected at the point (5, 2) then find its equation Sol. Let the equation of the line be  4x + 3y = 12 Find the equation of a line which passes through point A(2,3) and makes an angle of 45º with x-axis. If this line meet the line x + y + 1 = 0 at point P then distance AP Here x1 = 2, y1 = 3 and = 45º r = –3 2 x y  = 1 then a b the co–ordinates of point of intersection of this line and x axis and y axis are (a, 0) and (0, b) respectively (by putting x = 0 and y = 0) hence mid point of the intercept is (a/2, b/2)  a/2 = 5, b/2 = 2  a = 10, b = 4 hence required equation of the line is  2x + 5y = 20 Ex.10 Find the Length of perpendicular from the origin to the line 3x + 4y = 10 Sol. Converting the given equation into normal form by dividing 3x 4 y 10   5 5 5 32  4 2 = 5 is  3x 4 y  2 5 5 comparing with x cos  + y sin  = p p=2 Page # 4 STRAIGHT LINE Ex.11 Find the coordinates of a point which is at 3 distance from points (1,–3) of line 2x+3y+7=0 Sol. Slope of given line is = –  2 3 = tan–1(3/4) 2 3 tan = – Ex.15 If 3x + 4y –5 = 0 and 4x + ky – 8 = 0 are two perpendicular lines then k is - Hence 90º <  < 180º  2 sin = 13 (A) 3 13 Sol. 3 4 4 m2 = – k 3 4 ) × (– ) = –1 4 k k = –3 Ex.16 If 7x + 3y + 9 = 0 and y = kx + 7 are two desired point as 13 m1 = –  (– Putting r = 3 we get the co-ordinates of , (C) – 3 Two lines are perpendicular if m1m2 = –1 x 1 y3  r  3   2       13   13  9 (B) 4 (D) – 4 3 , cos = – Distance from of line 2x + 3y + 7 = 0 is x–1=– 1/ 2  2 B = tan–1 1  (1 / 2)(2)  parallel lines then find k y +3 = 6 Sol. 13 7 m1 = – m2 = k Two lines are parallel if m1=m2 3 k = – 7/3 or x = 1 – 9 13 , y = –3 + 6 Ex.17 Find the equation of straight lines which passes 13 through (3,4) and making an angle of 45º with Ex.12 Find the Normal form of line x + 3 y – 4 = 0 Sol. line x – y – 2 = 0 Sol. Given line x + 3 y = 4 Here m = 1 and tan = tan45º = 1 equation is y – 4 = (x – 3) i.e dividing both side by ( 3 ) 2  12 = 2 y – 4 = 0, x–3=0 Ex.18 One vertex of an equilateral triangle is (2,3) we get x 3y  =2 2 2 and the equation of line opposite to the vertex is x + y =2, then find the equation of remaining or x cos/3 + y sin /3 = 2 two sides Ex.13 Find the Angle between y = x+ 6 and y = 3 x+7 Sol. Here m1 = 1 tan= 1 3 1 3 m2 = 3 = tan–1 Sol. Slope of AB = m1 = – 1/2 Slope of BC = m2 = –2 Since the two sides make an angle of 60º each with side x + y = 2. Therefore equations of these sides will be 1 3 1 3 = 15º Ex.14 If A(1, 2), B(–1, 3) and C(3, –5) be vertices of a triangle then find B Sol. 1  tan 60º y – 3 = 1  (1) tan 60º (x – 2) =  1  3 1 3 (x – 2) y – 3 = (2 ± 3 )(x – 2)