Suppose A and B are the points at which the line x+y- =0 meets the pair of straight lines x^2+y^2-2 x-4 y+2=0 . If A O B=90^ , then a value of is

Step-by-step solution

Given, line is x+y- =0 x+y= x+y =1 ...(i) and pair of straight lines is x^2+y^2-2 x-4 y+2=0 We will make homogeneous of the above line. x^2+y^2-2 x( l )-4 y( l )+2( l )^2=0 x^2+y^2-2 x ( x+y )-4 y ( x+y )+2 ( x+y )^2=0 [using Eq. (i)] ^2 (x^2+y^2 )-2 x^2 -2 y x -4 x y -4 y^2 +2 (x^2+y^2+2 x y )=0 ( ^2-2 +2 ) x^2+(4 x y-6 x y ) ( ^2-4 +2 ) y^2=0 ( ^2-2 +2 ) x^2+(4-6 ) x y+ ( ^2-4 +2 ) y^2=0 Now, A O B=90^ where O is the origin. Condition for A D B=90^ is given by; ( . Coefficient of .x^2 )+ ( . Coefficient of .y^2 )