Question

If the points $P(12,8),Q(-2,a)$ and $R(6,0)$ are the vertices of a right angled triangle $PQR$, where $\angle R={90}^{\circ },$ then the value of $a$ is:

• A. $6$
• B. $-2$
• C. $-4$
• D. $-6$

Answer 1

Answer: (A)

$6$

$\Delta PQR$ is rt $\angle d$ at R
$\Rightarrow P{R}^2+Q{R}^2=P{Q}^2$
$\Rightarrow \{{(6-12)}^2+(0-8{)}^2\}+\{{(6+2)}^2+(0-a{)}^2\}$
$=\{{(-2-12)}^2+(a-8{)}^2\}$
$\Rightarrow (36+64)+(64+a^2)=(196+a^2-16a+64)$
$\Rightarrow a^2+164=a^2-16a+260$
$\Rightarrow 16a=260-164=96\Rightarrow a=6.$