The equations of sides $\mathbf{A B}, \mathbf{B C}$ and $\mathbf{C A}$ of a $\triangle A B C$ are $2 x+y=0,…

1. 50
2. 47
3. 65
4. 74

Solution

In $\triangle A B C$ Equation of $A B, B C$ and $C A$ are $2 x+y=0$, $x+P y=q$ and $x-y=3$ respectively. and $P(2,3)$ is orthocentre.
$\begin{aligned} & \text { Solving } \begin{aligned} & 2 x+y=0 \text { and } x-y=3 \\ & y=-2 x \text { and } x+2 x=3 \\ & \Rightarrow \quad 3 x=3 \\ & \Rightarrow \quad x=1, y=-2 \\ & \therefore A(1,-2\end{aligned} \\ & \text { Slope of } A P=\frac{3-(-2)}{2-1}=5 \\ & \text { Slope of } B C=-\frac{1}{p} \\ & A P \perp B C \\ & \Rightarrow \quad 5 \times\left(-\frac{1}{p}\right)=-1 \\ & p=5\end{aligned}$ Now for vertex $B$ $ \begin{aligned} & 2 x+y=0, y=-2 x \\ & x+5 y=q \\ & x-10 x=q \Rightarrow x=-\frac{q}{9}, y=\frac{2 q}{9} \\ & B \text { is }\left(-\frac{q}{9}, \frac{2 q}{9}\right) \\ & \text { Slope of } B P=\frac{\frac{2 q}{9}-3}{-\frac{q}{9}-2}=\frac{2 q-27}{-q-18} \\ & \end{aligned} $ Slope of $A C=1$ $ \begin{aligned} B P \perp A C \Rightarrow\left(\frac{q-27}{-q-18}\right)(1) & =-1 \\ \Rightarrow \quad 2 q-27 & =q+18 \\ 2 q-q & =18+27 \\ q & =45 \\ p+q & =5+45=50 \end{aligned} $

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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