Question

Let $a,b,c,d$ be non-zero numbers. If point of intersection of $6ax+2ay+c=0$ and $5bx+3by+d=0$ lies in second quadrant such that its distance from $x$-axis is twice its distance from $y$-axis then

• A. $7bc+10ad=0$
• B. $bc+2ad+=0$
• C. $7ad+10bc=0$
• D. $bc-2ad=0$

Answer 1

Answer: (B)

$bc+2ad+=0$

Let the coordinates of the intersection point in the second quadrant be $(-\alpha ,2\alpha )$ lies on both lines

$6ax+2ay+c=0$ and $5bx+3by+d=0$

The coordinates of the intersection point are represented as $(-\alpha ,2\alpha )$ because it is given that the distance from x-axis twice the distance from the y-axis

These coordinates should satisfy the equations of the intersecting lines

$\Rightarrow$ $6a(-\alpha )+2a\left(2\alpha \right)+c=0$

$\Rightarrow$ $-6a\alpha +4a\alpha +c=0$

$\Rightarrow$ $c=2a\alpha$

$\Rightarrow$ $\alpha =\frac{c}{2\alpha }....\left(1\right)$

From the other equation

$5b(-\alpha )+3b\left(2\alpha \right)+d=0$

$\Rightarrow$ $-5b\alpha +6b\alpha +d=0$

$\Rightarrow$ $b\alpha =-d$

$\alpha =-\frac{d}{a}.....\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$

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

$bc=-2ad$

$\Rightarrow$ $2ad+cb=0$

$\Rightarrow$ $bc+2ad=0$

$\therefore$ option b is the correct answer