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

Correct Answer: 32

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:

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