Let A(-2,-1),B(1,0),C( , ) and D( , ) be the vertices of a parallelogram ABCD.

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 \_\_\_\_\_\_\_\_\_\_\_.

Answer: 32

Accepted Answer

Since $ABCD$ is a parallelogram, $\vec{AB}=\vec{DC}$, i.e. $C-D = B-A = (3,1)$. With $C$ on $2x-y=5$ and $D$ on $3x-2y=6$, and $C=D+(3,1)$:
$$2(\gamma+3)-(\delta+1)=5 \implies 2\gamma-\delta=0$$
$$3\gamma-2\delta=6$$

Solving: $\delta=2\gamma$, so $3\gamma-4\gamma=6\implies\gamma=-6,\ \delta=-12$; then $\alpha=\gamma+3=-3,\ \beta=\delta+1=-11$.

$$|\alpha+\beta+\gamma+\delta| = |-3-11-6-12| = 32$$

Solutions · 1