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Coordinate Geometry
RD Sharma
CBSE
Grade 10

Question:

The fourth vertex $D$ of a parallelogram $ABCD$ whose three vertices are $A(-2, 3), B(6, 7)$ and $C(8, 3)$ is:
(a) $(0, -1)$
(b) $(0, 1)$
(c) $(1, 0)$
(d) $(-1, 0)$

Step-by-Step Solution

Key Concept: Diagonals of parallelogram bisect each other: $\dfrac{x_A + x_C}{2} = \dfrac{x_B + x_D}{2} \Rightarrow -2 + 8 = 6 + x_D \Rightarrow x_D = 0$. $\dfrac{y_A + y_C}{2} = \dfrac{y_B + y_D}{2} \Rightarrow 3 + 3 = 7 + y_D \Rightarrow y_D = -1$. $D(0, -1)$.
$-2 + 8 = 6 + x_D \Rightarrow x_D = 0$. $3 + 3 = 7 + y_D \Rightarrow y_D = -1$. $D(0, -1)$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Using diagonal midpoint property to find $D(0, -1)$: 1.0 Mark

Correct Answer: $(0, -1)$
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