Question

Consider a triangle $PQR$ in which the relation $Q{R}^2+P{R}^2=5P{Q}^2$ holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then $\angle QGM$ is always

• A. less than ${45}^{\circ }$
• B. obtuse
• C. a right angle
• D. acute and larger than ${45}^{\circ }$

Answer 1

The correct option is C a right angle


Given :
$Q{R}^2+P{R}^2=5P{Q}^2$
$\therefore (2a{)}^2+[(x+a{)}^2+y^2]=5\left[\right(x-a{)}^2+y^2]$

$\Rightarrow 4a^2+x^2+a^2+2ax+y^2=5(x^2+a^2-2ax+y^2)$

$\Rightarrow 4a^2+x^2+a^2-5a^2+10ax-5x^2+2ax=5y^2-y^2$

$\Rightarrow 4y^2=12ax-4a^2$

$\Rightarrow y^2=3ax-x^2=x(3a-x)$.........(i)

Now as $G$ is the point of intersection of medians,
$G=(\frac{x+a-a}{3},\frac{y+0+0}{3})=(\frac{x}{3},\frac{y}{3})$
Slope of $QG$,
$m_{QG}=\frac{0-\frac{y}{3}}{a-\frac{x}{3}}=\frac{y}{x-3a}$
Slope of $MG$,
$m_{MG}=\frac{0-\frac{y}{3}}{0-\frac{x}{3}}=\frac{y}{x}$
$m_{QG}\times m_{MG}=\frac{y}{x-3a}\times \frac{y}{x}=\frac{y^2}{x(x-3a)}$
$=-1$

Hence the $\angle QGM$ is always right angle.