Question

The coordinates of points $P,Q,R$ and $S$ are $(-3,5),(4,-2),(p,3p)$ and $(6,3)$ respectively, and the area of $△PQR$ and $△QRS$ are in ratio $2:3,$ then possible value of $p$ can be

• A. $-\frac{5}{43}$
• B. $\frac{3}{43}$
• C. $\frac{46}{41}$
• D. $\frac{45}{41}$

Answer 1

The correct option is D $\frac{45}{41}$

Given, $P\equiv (-3,5),Q\equiv (4,-2),R\equiv (p,3p)$ and $S\equiv (6,3)$

Now, area of $△PQR$
$=\frac{1}{2}|-3(-2-3p)+4(3p-5)+p(5+2)|=\frac{1}{2}|28p-14|$

And, area of $△QRS$
$=\frac{1}{2}|4(3p-3)+p(3+2)+6(-2-3p)|=\frac{1}{2}|p+24|$

Given, area of $△PQR$ and $△QRS$ are in ratio $2:3.$
$\Rightarrow \left|\frac{28p-14}{p+24}\right|=\frac{2}{3}\Rightarrow \frac{28p-14}{p+24}=\pm \frac{2}{3}$

When $\frac{28p-14}{p+24}=\frac{2}{3}$,
$84p-42=2p+48\Rightarrow p=\frac{90}{82}\Rightarrow p=\frac{45}{41}$

When $\frac{28p-14}{p+24}=-\frac{2}{3}$,
$84p-42=-2p-48\Rightarrow p=-\frac{6}{86}\Rightarrow p=-\frac{3}{43}$