Straight Lines
Parallelogram with Vertices on Given Lines — |α+β+γ+δ|
nta_pyq_2024_jan
Grade 11
Question:
Let $A(-2,-1)$, $B(1,0)$, $C(\alpha,\beta)$ and $D(\gamma,\delta)$ be the vertices of a parallelogram $ABCD$. If the point $C$ lies on $2x-y=5$ and the point $D$ lies on $3x-2y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to
Step-by-Step Solution
Key Concept: Diagonals bisect each other: midpoint of $AC$ = midpoint of $BD$. $\frac{\alpha-2}{2}=\frac{\gamma+1}{2}$ and $\frac{\beta-1}{2}=\frac{\delta+0}{2}$. With constraints $2\alpha-\beta=5$ and $3\gamma-2\delta=6$, solve the system.
$\alpha=-3,\beta=-11,\gamma=-6,\delta=-12$. $|\alpha+\beta+\gamma+\delta|=32$.
Correct Answer: 32