Question

The line parallel to the X–axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0 where (a, b) ≠ (0, 0) is:

(A) above the X-axis at a distance of (2/3) from it

(B) above the X-axis at a distance of (3/2) from it 

(C) below the X-axis at a distance of (2/3) from it

(D) below the X-axis at a distance of (3/2) from it

Answer 1

The correct option (D) below the X-axis at a distance of (3/2) from it

Explanation:

y coordinate of point of intersection of lines ax + 2by + 3b = 0 and

bx – 2ay – 3a = 0 is given by,

put x = [(– 3b – 2by)/a] in bx – 2ay – 3a = 0, we get,

(b/a)(– 3b – 2by) – 2ay – 3a = 0

∴ [(– 3b²)/a] – [(2b²)/a]y – 2ay – 3a = 0

∴ y[2a + (2b²/a)] = [(– 3b²/a) – 3a

∴ y[(2a² + 2b²)/a] = [(– 3b² – 3a²)/a]

∴ y(2) = – 3 

i.e.   y = [(– 3)/2]

lines is parallel to x-axis hence equation is y = [(– 3)/2]

∴ line is at distance (3/2).