Straight Lines Question 14
Question 14 - 2024 (31 Jan Shift 2)
Let $A(-2,-1), B(1,0), C(\alpha, \beta)$ and $D(\gamma, \delta)$ be the vertices of a parallelogram $A B C D$. If the point $C$ lies on $2 x-y=5$ and the point $D$ lies on $3 x-2 y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to
Show Answer
Answer (32)
Solution
$\mathrm{P} \equiv\left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) \equiv\left(\frac{\gamma+1}{2}, \frac{\delta}{2}\right)$
$\frac{\alpha-2}{2}=\frac{\gamma+1}{2}$ and $\frac{\beta-1}{2}=\frac{\delta}{2}$
$\Rightarrow \alpha-\gamma=3 \ldots(1), \quad \beta-\delta=1$
Also, $(\gamma, \delta)$ lies on $3 x-2 y=6$
$3 \gamma-2 \delta=6$
and $(\alpha, \beta)$ lies on $2 x-y=5$
$\Rightarrow 2 \alpha-\beta=5$.
Solving (1), (2), (3), (4)
$\alpha=-3, \beta=-11, \gamma=-6, \delta=-12$
$|\alpha+\beta+\gamma+\delta|=32$
