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If the point of intersection of the lines $2 a x+4 a y+c=0$ and $7 b x+3 b y-d=0$ lies in the 4 th quadrant and is equidistant from the two axes, where $a, b, c$ and $d$ are non-zero numbers, then $a d: b c$ equals to
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Verified Answer
The correct answer is:
$2: 1$
Let coordinate of the point be $(\alpha,-\alpha)$
since, $(\alpha,-\alpha)$ lie on $2 a x+4 a y+c=0$
and $\quad 7 b x+3 b y-d=0$
$\therefore \quad 2 a \alpha-4 a \alpha+c=0$
$\Rightarrow \quad-2 a \alpha+c=0$
$\Rightarrow \quad \alpha=\frac{c}{2 a}$
Also, $7 b \alpha-3 b \alpha-d=0$
$\Rightarrow \quad 4 b \alpha-d=0$
$\Rightarrow \quad \alpha=\frac{d}{4 b}$
From Eqs. (i) and (ii),
$\begin{aligned} \frac{c}{2 a} &=\frac{d}{4 b} \\
\Rightarrow 2 a d &=4 b c \end{aligned}$
$\Rightarrow \quad \frac{a d}{b c}=\frac{4}{2}$
$\Rightarrow \quad \frac{a d}{b c}=\frac{2}{1}$
$a d: b c=2: 1$
since, $(\alpha,-\alpha)$ lie on $2 a x+4 a y+c=0$
and $\quad 7 b x+3 b y-d=0$
$\therefore \quad 2 a \alpha-4 a \alpha+c=0$
$\Rightarrow \quad-2 a \alpha+c=0$
$\Rightarrow \quad \alpha=\frac{c}{2 a}$
Also, $7 b \alpha-3 b \alpha-d=0$
$\Rightarrow \quad 4 b \alpha-d=0$
$\Rightarrow \quad \alpha=\frac{d}{4 b}$
From Eqs. (i) and (ii),
$\begin{aligned} \frac{c}{2 a} &=\frac{d}{4 b} \\
\Rightarrow 2 a d &=4 b c \end{aligned}$
$\Rightarrow \quad \frac{a d}{b c}=\frac{4}{2}$
$\Rightarrow \quad \frac{a d}{b c}=\frac{2}{1}$
$a d: b c=2: 1$
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