Q83 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ and $\mathrm{d}$ be non-zero numbers. If the point of intersection of the lines $4 a x+2 a y+c=0$ and $5 b x+2 b y$ $+\mathrm{d}=0$ lies in the fourth quadrant and is equidistant from the two coordinate axes, then
- Correct answerA$3 \mathrm{bc}-2 \mathrm{ad}=0$
- B$3 \mathrm{bc}+2 \mathrm{ad}=0$
- C$2 \mathrm{bc}-3 \mathrm{ad}=0$
- D$2 \mathrm{bc}+3 \mathrm{ad}=0$
Solution
Given: $4 \mathrm{ax}+2 \mathrm{ay}+\mathrm{c}=0$
$5 b x+2 b y+d=0$
Eqn. (i) $\times b-$ (ii) $\times a$
$$
-a b x+b c-a d=0
$$
$$
\Rightarrow \mathrm{x}=\frac{\mathrm{bc}-\mathrm{ad}}{\mathrm{ab}}
$$
Putting the value of $x$ in eqn. (ii), we get
$$
\begin{aligned}
& 5 b\left(\frac{b c-a d}{a b}\right)+2 b y+d=0 \\
& \Rightarrow y=\frac{4 a d-5 b c}{2 a b}
\end{aligned}
$$
Point of intersection is $P\left(\frac{b c-a d}{a b}, \frac{4 a d-5 b c}{2 a b}\right)$
$\because \mathrm{P}$ is in fourth quadrant and equidistant from the two coordinate axes.
$$
\begin{aligned}
& \therefore-\frac{4 a d-5 b c}{2 a b}=\frac{b c-a d}{a b} \\
& \Rightarrow-4 a d+5 b c=2 b c-2 a d \\
& \Rightarrow 3 b c=2 a b \Rightarrow 3 b c-2 a d=0
\end{aligned}
$$
Hence, the correct answer is option: 'A'.