JEE-Mains 2014 (19/30)

Geometry Level 3

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:

3bc-2ad=0

3bc+2ad=0

2bc+3ad=0

2bc-3ad=0

1 solution

Akhil Bansal
Jun 27, 2015

Let both the equations meets at point (k,-k)
It means these point will satisfy the given equation.
$\Rightarrow 4ak-2ak+c=0 \rightarrow 2ak=-c$

$\Rightarrow 5bk-2bk+d=0 \rightarrow 3bk=-d$

Divide both the equations and cross multiply them,
$\Rightarrow \boxed{3bc-2ad=0}$ is the solution.

JEE-Mains 2014 (19/30)

1 solution

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