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 1

Given that $A(-2, -1)$ and $B(1, 0)$ are two vertices of the parallelogram and $C(\alpha, \beta)$ and $D(\gamma, \delta)$ are the other two vertices.

Since $P$ is the midpoint of diagonals $AC$ and $BD$, we have:

$P = \left(\frac{\alpha - 2}{2}, \frac{\beta - 1}{2}\right) = \left(\frac{\gamma + 1}{2}, \frac{\delta}{2}\right)$

Equating coordinates:

$\frac{\alpha - 2}{2} = \frac{\gamma + 1}{2} \quad \text{and} \quad \frac{\beta - 1}{2} = \frac{\delta}{2}$

Simplifying:

$\alpha - 2 = \gamma + 1 \implies \alpha - \gamma = 3 \quad (1)$ $\beta - 1 = \delta \implies \beta - \delta = 1 \quad (2)$

Given that $(\gamma, \delta)$ lies on the line $3x - 2y = 6$:

$3\gamma - 2\delta = 6 \quad (3)$

Also, $(\alpha, \beta)$ lies on the line $2x - y = 5$:

$2\alpha - \beta = 5 \quad (4)$

Solving equations (1), (2), (3), and (4) simultaneously: From (1) and (2):

$\alpha = \gamma + 3, \quad \beta = \delta + 1$

Substitute these values into (3) and (4):

$3\gamma - 2\delta = 6$ $2(\gamma + 3) - (\delta + 1) = 5$

Simplifying:

$3\gamma - 2\delta = 6$ $2\gamma + 6 - \delta - 1 = 5 \implies 2\gamma - \delta = 0$

Solving these equations:

$\gamma = -6, \quad \delta = -12, \quad \alpha = -3, \quad \beta = -11$

Thus, the value of $|\alpha + \beta + \gamma + \delta|$ is:

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