Let PQR be a right angled isosceles triangle, | vedclass

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Question

Let the slopes of $P Q$ and $P R$ be $m$ and $-\frac{1}{m}$ respectively. since $	riangle P Q R$ is an isosceles triangle, $\angle P Q R=\angle P R Q

Vedclass · Accepted Answer

$3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$

Vedclass · Answer

Let the slopes of $P Q$ and $P R$ be $m$ and $-\frac{1}{m}$ respectively. since $	riangle P Q R$ is an isosceles triangle, $\angle P Q R=\angle P R Q$ $\Rightarrow\left|\frac{m+2}{1-2 m}ight|=\left|\frac{-\frac{1}{m}+2}{1+\frac{2}{m}}ight|$ 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 $y-1=\left(-\frac{1}{3}ight)(x-2)$

Thus, joint equation representing $P Q$ and $P R$ is

$(3(x-2)-(y-1))((x-2)+3(y-1))=0$

$\Rightarrow 3(x-2)^{2}-3(y-1)^{2}+8(x-2)(y-1)=0$

$\Rightarrow 3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25=0$

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

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