- Home
- Standard 11
- Mathematics
Let $PQR$ be a right angled isosceles triangle, right angled at $P\, (2, 1)$. If the equation of the line $QR$ is $2x + y = 3$, then the equation representing the pair of lines $PQ$ and $PR$ is
$3x^2 - 3y^2 + 8xy + 20x + 10y + 25 = 0$
$3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$
$3x^2 - 3y^2 + 8xy + 10x + 15y + 20 = 0$
$3x^2 - 3y^2 - 8xy - 10x - 15y - 20 = 0$
Solution
Let the slopes of $P Q$ and $P R$ be $m$ and $-\frac{1}{m}$ respectively. since $\triangle P Q R$ is an isosceles triangle, $\angle P Q R=\angle P R Q$ $\Rightarrow\left|\frac{m+2}{1-2 m}\right|=\left|\frac{-\frac{1}{m}+2}{1+\frac{2}{m}}\right|$ as slope of $Q R=2$
$\Rightarrow m+2=\pm(1-2 m) \Rightarrow m=3$ or $-\frac{1}{3}$
So the equation of $P Q$ and $P R$ are $(y-1)=3(x-2)$ and $y-1=\left(-\frac{1}{3}\right)(x-2)$
Thus, joint equation representing $P Q$ and $P R$ is
$(3(x-2)-(y-1))((x-2)+3(y-1))=0$
$\Rightarrow 3(x-2)^{2}-3(y-1)^{2}+8(x-2)(y-1)=0$
$\Rightarrow 3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0$
