JEE Main 2014 · 6 Apr · Shift 1 · Mathematics Q79

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MathematicsSingle correct+4 / -1Question 79 of 90

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

Answer and worked solution

Official answer: A

The intersection is $(k,-k)$ with $k>0$ (fourth quadrant, equidistant from the axes).

From the first line: $4ak-2ak+c=0$ , so $k=-\frac c{2a}$ .
From the second: $5bk-2bk+d=0$ , so $k=-\frac d{3b}$ .

Equating, $3bc=2ad$ , i.e. $3bc-2ad=0$ .

Topics:Equations of a line

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JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 79