Q76 / 155 · WBJEE 2019 · Mathematics

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

  • A
    $2: 3$
  • Correct answer
    B
    $2: 1$
  • C
    $1: 1$
  • D
    $3: 2$

Solution

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), $\begin{aligned} \frac{c}{2 a} &=\frac{d}{4 b} \\ \Rightarrow 2 a d &=4 b c \end{aligned}$ $\Rightarrow \quad \frac{a d}{b c}=\frac{4}{2}$ $\Rightarrow \quad \frac{a d}{b c}=\frac{2}{1}$ $a d: b c=2: 1$

Hence, the correct answer is option: 'B'.

Relevant Questions