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\( \_\_\_\_\_\).
Correct Answer: 32
Approach Solution - 1
Given:
\[ P \equiv \left( \frac{\alpha - 2}{2}, \frac{\beta - 1}{2} \right) \equiv \left( \frac{\gamma + 1}{2}, \frac{\delta}{2} \right) \]
\[ \frac{\alpha - 2}{2} = \frac{\gamma + 1}{2} \quad \text{and} \quad \frac{\beta - 1}{2} = \frac{\delta}{2} \]
\[ \Rightarrow \alpha - \gamma = 3 \quad \text{....(1)} \] \[ \beta - \delta = 1 \quad \text{....(2)} \]
Step 1: Point \((\gamma, \delta)\) lies on the line \(3x - 2y = 6\)
\[ 3\gamma - 2\delta = 6 \quad \text{....(3)} \]
Step 2: Point \((\alpha, \beta)\) lies on the line \(2x - y = 5\)
\[ 2\alpha - \beta = 5 \quad \text{....(4)} \]
Step 3: Solving equations (1), (2), (3), and (4)
\[ \alpha = -3, \quad \beta = -11, \quad \gamma = -6, \quad \delta = -12 \]
Step 4:
\[ |\alpha + \beta + \gamma + \delta| = |-3 - 11 - 6 - 12| = 32 \]
Final Answer:
\[ \boxed{32} \]
Approach Solution -2
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:
\[ |\alpha + \beta + \gamma + \delta| = | -3 + (-11) + (-6) + (-12)| = | -32| = 32 \]
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