Question

The minimum value of $p$ for which the lines $3x-4y=2,3x-4y=12,12x+5y=7$ and $12x+5y=p$ constitute the sides of a rhombus is

• A. $33$
• B. $19$
• C. $-19$
• D. $9$

Answer 1

Answer: (C)

If the parallelogram is a rhombus, the distance between pair of parallel sides is equal.
Hence, $\frac{12-2}{\sqrt{3^2+4^2}}=\pm \frac{p-7}{\sqrt{12^2+5^2}}$
$\Rightarrow p-7=\pm 26$
$\Rightarrow p=33$ or $-19$
Hence, the minimum value of $p$ is $-19$