Let $\mathrm{PQR}$ be a right angled isosceles triangle, right angled at $\mathrm{P}(2,1)$. If the equation…

Let $\mathrm{PQR}$ be a right angled isosceles triangle, right angled at $\mathrm{P}(2,1)$. If the equation of the line $\mathrm{QR}$ is $2 x+y=3$, then the equation representing the pair of lines $P Q$ and $P R$ is
  1. $3 x^2-3 y^2+8 x y+20 x+10 y+25=0$
  2. $3 x^2-3 y^2+8 x y-20 x-10 y+25=0$
  3. $3 x^2-3 y^2+8 x y+10 x+15 y+20=0$
  4. $3 x^2-3 y^2-8 x y-10 x-15 y-20=0$

Solution

$\begin{array}{ll} & \text { Slope of } Q R=-2 . \\ & \text { Slope of } P Q=\mathrm{m}_1 \\ \therefore \quad & \tan 45^{\circ}=\left|\frac{\mathrm{m}_1+2}{1+\mathrm{m}_1(-2)}\right| \\ \Rightarrow & 1=\left|\frac{\mathrm{m}_1+2}{1-2 \mathrm{~m}_1}\right| \\ \Rightarrow & \mathrm{m}_1=-\frac{1}{3}\end{array}$ $\therefore \quad$ Equation of $\mathrm{PQ}$ passing through point $\mathrm{P}(2,1)$ and having slope $\frac{-1}{3}$ is $\begin{aligned} & y-1=-\frac{1}{3}(x-2) \\ & \Rightarrow 3(y-1)+(x-2)=0 \end{aligned}$ Slope of $\mathrm{PR}=\mathrm{m}_2=3$ $\therefore \quad$ equation of $\mathrm{PR}$ is $\begin{aligned} & y-1=3(x-2) \\ & \Rightarrow(y-1)-3(x-2)=0 \end{aligned}$ $\therefore \quad$ The joint equation of the lines is $\begin{aligned} & {[3(y-1)+(x-2)][(y-1)-3(x-2)]=0} \\ & \Rightarrow 3(y-1)^2-8(y-1)(x-2)-3(x-2)^2=0 \\ & \Rightarrow 3\left(x^2-4 x+4\right)+8(x y-x-2 y+2) \\ & \quad-3\left(y^2-2 y+1\right)=0 \\ & \Rightarrow 3 x^2-3 y^2+8 x y-20 x-10 y+25=0 \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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