Let [, , , Z] and let [A(, ), B(1,0), C(, )] and D(1,2) be the vertices of a parallelogram…

Question

Let $$\alpha, \beta, \gamma, \delta \in \mathbb{Z}$$ and let $$A(\alpha, \beta), B(1,0), C(\gamma, \delta)$$ and $D(1,2)$ be the vertices of a parallelogram $$\mathrm{ABCD}$$. If $$A B=\sqrt{10}$$ and the points $$\mathrm{A}$$ and $$\mathrm{C}$$ lie on the line $3 y=2 x+1$, then $$2(\alpha+\beta+\gamma+\delta)$$ is equal to

Accepted Answer

Let E is mid point of diagonals

$$\frac{\alpha+\gamma}{2}=\frac{1+1}{2}$$

$$\alpha+\gamma=2$$

\& $$\frac{\beta+\delta}{2}=\frac{2+0}{2}$$

$$\beta+\delta=2$$

$$2(\alpha+\beta+\gamma+\delta)=2(2+2)=8$$

Solutions · 1