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Coordinate Geometry
NCERT Exemplar
CBSE
Grade 10
Question:
If $A(-5, 7), B(-4, -5), C(-1, -6)$ and $D(4, 5)$ are the vertices of a quadrilateral $ABCD$, prove using distance formula whether $ABCD$ is a parallelogram or not.
Step-by-Step Solution
Key Concept: Calculate side lengths or check whether midpoints of diagonals $AC$ and $BD$ coincide.
Midpoint of diagonal $AC = \left(\dfrac{-5 + (-1)}{2}, \dfrac{7 + (-6)}{2}\right) = \left(\dfrac{-6}{2}, \dfrac{1}{2}\right) = \left(-3, \dfrac{1}{2}\right)$. [1.0 Mark] Midpoint of diagonal $BD = \left(\dfrac{-4 + 4}{2}, \dfrac{-5 + 5}{2}\right) = (0, 0)$. [1.0 Mark] Since the midpoints of diagonals $AC$ and $BD$ do not coincide $\left(\left(-3, \dfrac{1}{2}\right) eq (0, 0)\right)$, the diagonals do not bisect each other. Hence, $ABCD$ is NOT a parallelogram. [1.0 Mark]
--- 🎯 Official CBSE Marking Scheme: Calculating midpoint of $AC = (-3, 1/2)$: 1.0 Mark Calculating midpoint of $BD = (0, 0)$: 1.0 Mark Comparing midpoints and concluding NOT a parallelogram: 1.0 Mark
Correct Answer:
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