Question

Consider the following two families of lines represented by the equations (x - y -  6) + λ(2x + y - 3) = 0 and (x + 2y -  4) μ(3x -  2y - 4) = 0. If these families of lines are at right angles to each other, then their points of intersection lie on the curve

(A)  x² -  y² + 3x - 4y - 3 = 0

(B)  x² + y² - 5x + 2y + 3 = 0 

(C)  x² +  y² + 3x + 4y - 3 = 0 

(D)  x² -  y² - 3x + 4y - 3 = 0 

Answer 1

Correct option  B)  x² + y² - 5x + 2y + 3 = 0 

Explanation :

the equation (x − y − 6) + λ(2x + y − 3) = 0 represents the family of lines passing through the intersection of the lines x − y− 6 = 0 and 2x + y − 6 = 0 which is (3, −3). Similarly, the second equation represents a family of concurrent lines which is concurrent at the point of intersection of the lines x + 2y −4 = 0 and 3x − 2y − 4 = 0 which is (2, 1). Suppose the line through the point (3, - 3) is

y + 3 = m (x + 3)   ...(1)

Then the line through the point (2, 1) and perpendicular to the line given in eq.(1) is

y - 1 =  - 1/m(x - 2)   ...(2)

From Eqs. (1) and (2), we have