Straight Line And Pair Of Straight Lines Ques 14
- Let $a, b, c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4 a x+2 a y+c=0$ and $5 b x+2 b y+d=0$ lies in the fourth quadrant and is equidistant from the two axes, then
(2014 Main)
(a) $2 b c-3 a d=0$
(b) $2 b c+3 a d=0$
(c) $2 a d-3 b c=0$
(d) $3 b c+2 a d=0$
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Solution:
Formula:
- Let coordinate of the intersection point in fourth quadrant be $(\alpha,-\alpha)$.
Since, $(\alpha,-\alpha)$ lies on both lines $4 a x+2 a y+c=0$ and $5 b x+2 b y+d=0$.
$$ \begin{aligned} & \therefore \quad 4 a \alpha-2 a \alpha+c=0 \Rightarrow \alpha=\frac{-c}{2 a} \\ & \text { and } \quad 5 b \alpha-2 b \alpha+d=0 \Rightarrow \alpha=\frac{-d}{3 b} \end{aligned} $$
From Eqs. (i) and (ii), we get
$$ \begin{aligned} \frac{-c}{2 a} & =\frac{-d}{3 b} \Rightarrow 3 b c=2 a d \\ \Rightarrow \quad 2 a d-3 b c & =0 \end{aligned} $$
BETA
