If $A(-5, 7), B(-4, -5), C(-1, -6)$ and $D(4, 5)$ are the vertices of a quadrilateral $ABCD$, prove that the midpoints of its sides form a parallelogram.
Step-by-Step Solution
Key Concept: Midpoints: $P(AB) = (-4.5, 1)$, $Q(BC) = (-2.5, -5.5)$, $R(CD) = (1.5, -0.5)$, $S(DA) = (-0.5, 6)$. Midpoint of $PR = (-1.5, 0.25)$. Midpoint of $QS = (-1.5, 0.25)$. Since diagonal midpoints coincide, $PQRS$ is a parallelogram.
Midpoints: $P = (-4.5, 1), Q = (-2.5, -5.5), R = (1.5, -0.5), S = (-0.5, 6)$. [1.5 Marks]
Midpoint of $PR = \left(\dfrac{-4.5 + 1.5}{2}, \dfrac{1 - 0.5}{2}\right) = (-1.5, 0.25)$. [0.5 Mark]
Midpoint of $QS = \left(\dfrac{-2.5 - 0.5}{2}, \dfrac{-5.5 + 6}{2}\right) = (-1.5, 0.25)$. Since diagonal midpoints coincide, $PQRS$ is a parallelogram. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding midpoints $P, Q, R, S$: 1.5 Marks
Showing diagonal midpoints of $PQRS$ coincide: 1.5 Marks
Correct Answer: