Straight Lines Question 2
Question 2 - 24 January - Shift 2
The equations of the sides $AB, BC$ and $CA$ of $a$ triangle $A B C$ are: $2 x+y=0, x+p y=21 a,(a \neq 0)$ and $x-y=3$ respectively. Let $P(2, a)$ be the centroid of $\triangle ABC$. Then $(BC)^{2}$ is equal to
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Answer: 122)
Solution:
Formula: Distance Formula, Centroid Formula
Assume B($\alpha$,$-2 \alpha$) and C($\beta+3$, $\beta$)
$$ \begin{aligned} & \frac{\alpha+\beta+3+1}{3}=2 \text { also } \frac{-2 \alpha-2+\beta}{3}=a \ & \Rightarrow \alpha+\beta=2 \quad-2 \alpha-2+\beta=3 a \ & \Rightarrow \beta=2-\alpha-2 \alpha-\not 22 \not 2-\alpha=3 a \Rightarrow \alpha=-a \end{aligned} $$
Now both $B$ and $C$ lies as given line $$ \begin{aligned} & \alpha-p \cdot 2 \alpha=21 a \ & \alpha(1-2 p)=21 a \ldots(1) \ & -\alpha(1-2 p)=21 a \Rightarrow p=11 \ & \beta+3+p \beta=21 a \ & \beta+3+11 \beta=21 a \ & 21 \alpha+12 \beta+3=0 \end{aligned} $$
Also $\beta=2-\alpha$ $$ \begin{aligned} & 21 \alpha+12(2-\alpha)+3=0 \ & 21 \alpha+24-12 \alpha+3=0 \ & 9 \alpha+27=0 \ & \alpha=-3, \beta=5 \end{aligned} $$
So $B C=\sqrt{122}$ and $(B C)^2=122$
