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)

$3bc-2ad=0$

Since point of intersection lies in the fourth quadrant and is equidistant from coordinate axes,

the $x$ and $y$ co-ordinates will be same

Hence the coordinates become $(h,-h)$

Passing $4ax+2ay+c=0$ through $(h,-h)$

$\Rightarrow 4ah-2ah+c=0$

$\Rightarrow h=-\frac{c}{2a}----------\left(1\right)$

Also passing the second line $5bx+2by+d=0$ through $(h,-h)$

$\Rightarrow 5bh-2bh+d=0$

$\Rightarrow h=-\frac{d}{3b}----\left(2\right)$

From eq (1) and (2)

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

$\Rightarrow 3bc-2ad=0$