Let A(-2,-1), B(1,0), C(\alpha, \beta) and D(\gamma, \delta) be the vertices of a parallelogram A B C D . If the point C lies on 2 x-y=5 and the point D lies on 3 x-2 y=6 , then the value of |\alpha+\beta+\gamma+\delta| is equal to \underline\hspace2cm .

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q54 with Solution

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Correct answer: 32

Use the parallelogram property

For parallelogram $ABCD$ with vertices in order, the diagonals bisect each other, so $A + C = B + D$ coordinate-wise.

Given:

\quad B(1,0), \quad C(\alpha,\beta), \quad D(\gamma,\delta)$$ we get $$(-2,-1) + (\alpha,\beta) = (1,0) + (\gamma,\delta).$$ Thus, $$\alpha - \gamma = 3, \qquad \beta - \delta = 1.$$ So, $$\alpha = \gamma + 3, \qquad \beta = \delta + 1.$$ --- 2. **Use the condition that $C$ lies on $2x-y=5$** Since $C(\alpha,\beta)$ lies on $2x-y=5$, $$2\alpha - \beta = 5.$$ Substitute $\alpha = \gamma+3$ and $\beta = \delta+1$: $$2(\gamma+3) - (\delta+1) = 5$$ $$2\gamma + 6 - \delta - 1 = 5$$ $$2\gamma - \delta = 0.$$ So, $$\delta = 2\gamma.$$ --- 3. **Use the condition that $D$ lies on $3x-2y=6$** Since $D(\gamma,\delta)$ lies on $3x-2y=6$, $$3\gamma - 2\delta = 6.$$ Using $\delta = 2\gamma$: $$3\gamma - 2(2\gamma) = 6$$ $$3\gamma - 4\gamma = 6$$ $$-\gamma = 6$$ $$\gamma = -6.$$ Hence, $$\delta = 2\gamma = -12.$$ Then, $$\alpha = \gamma + 3 = -6 + 3 = -3,$$ $$\beta = \delta + 1 = -12 + 1 = -11.$$ --- 4. **Compute the required value** $$\alpha + \beta + \gamma + \delta = -3 + (-11) + (-6) + (-12) = -32.$$ Therefore, $$|\alpha+\beta+\gamma+\delta| = |-32| = 32.$$ --- 5. **Compare with stored answer** Derived answer = $32$. Stored correct answer = $32$. They agree.

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View written solutionFree

Correct answer: 32

Use the parallelogram property

For parallelogram $ABCD$ with vertices in order, the diagonals bisect each other, so $A + C = B + D$ coordinate-wise.

Given:

\quad B(1,0), \quad C(\alpha,\beta), \quad D(\gamma,\delta)$$ we get $$(-2,-1) + (\alpha,\beta) = (1,0) + (\gamma,\delta).$$ Thus, $$\alpha - \gamma = 3, \qquad \beta - \delta = 1.$$ So, $$\alpha = \gamma + 3, \qquad \beta = \delta + 1.$$ --- 2. **Use the condition that $C$ lies on $2x-y=5$** Since $C(\alpha,\beta)$ lies on $2x-y=5$, $$2\alpha - \beta = 5.$$ Substitute $\alpha = \gamma+3$ and $\beta = \delta+1$: $$2(\gamma+3) - (\delta+1) = 5$$ $$2\gamma + 6 - \delta - 1 = 5$$ $$2\gamma - \delta = 0.$$ So, $$\delta = 2\gamma.$$ --- 3. **Use the condition that $D$ lies on $3x-2y=6$** Since $D(\gamma,\delta)$ lies on $3x-2y=6$, $$3\gamma - 2\delta = 6.$$ Using $\delta = 2\gamma$: $$3\gamma - 2(2\gamma) = 6$$ $$3\gamma - 4\gamma = 6$$ $$-\gamma = 6$$ $$\gamma = -6.$$ Hence, $$\delta = 2\gamma = -12.$$ Then, $$\alpha = \gamma + 3 = -6 + 3 = -3,$$ $$\beta = \delta + 1 = -12 + 1 = -11.$$ --- 4. **Compute the required value** $$\alpha + \beta + \gamma + \delta = -3 + (-11) + (-6) + (-12) = -32.$$ Therefore, $$|\alpha+\beta+\gamma+\delta| = |-32| = 32.$$ --- 5. **Compare with stored answer** Derived answer = $32$. Stored correct answer = $32$. They agree.

Straight Lines and Pair of Straight Lines question

Straight Lines and Pair of Straight Lines question

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