Question

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes then

• A. $2bc-3ad=0$
• B. $2bc+3ad=0$
• C. $3bc-2ad=0$
• D. $3bc+2ad=0$

Answer 1

Answer: (C)

$3bc-2ad=0$

Solve the equations of two lines to get a point and satisfy it in the third equation of a line.

$4ax+2ay+c=0$ and $5bx+2by+d=0$

Let the point of intersection be $(k,-k)$

From both equations, we get

$4ak-2ak+c=0,5bk-2bk+d=0$

$2ak+c=0,3bk+d=0$

$k=\frac{-c}{2a},k=\frac{-d}{3b}$

$\frac{-c}{2a}=\frac{-d}{3b}$

$\therefore 3bc-2ad=0$

Option C