Question

The line parallel to the X-axis and passing through the point of intersection of the lin `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 equation of a line passing through the intersection of given lines is
(ax + 2by + 3b) + `lambda` ( bx - 2ay - 3a) = 0
or `x (a + lambdab) + y (2b - 2 lambda a) + (3b - 3 lambda a) = 0 " " … (i)`
`therefore ` Slope = 0 `implies a + lambda b = 0 implies = - (a)/(b) `
Putting `lambda = - (a)/(b)` in (i) , we obtain
`(2)/(b) (a^(2) + b^(2)) y + (3)/(b) (a^(2) + b^(2)) = 0 implies 2y + 3 = 0 implies y = - (3)/(2)`
Clearly , it is below the x-axis at a distance of `(3)/(2)` units from x-axis .