The fourth vertex $D$ of a parallelogram $ABCD$ whose three vertices are $A(-2, 3)$, $B(6, 7)$ and $C(8, 3)$…

Question

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

Accepted Answer

## Model Answer **Option (B) $(0, -1)$** In a parallelogram, diagonals bisect each other, so mid-point of AC = mid-point of BD. Mid-point of AC $= \left(\dfrac{-2+8}{2},\ \dfrac{3+3}{2}\right) = (3, 3)$ Let $D = (x, y)$. Then mid-point of BD $= \left(\dfrac{6+x}{2},\ \dfrac{7+y}{2}\right) = (3, 3)$ $\Rightarrow 6+x = 6 \Rightarrow x = 0$ and $7+y = 6 \Rightarrow y = -1$ Therefore, $D = (0, -1)$. ## Explanation The key property used is: **diagonals of a parallelogram bisect each other**, so mid-point of diagonal AC = mid-point of diagonal BD (as in Example 10 of Section 7.3). Equate coordinates separately to find $x$ and $y$. This is a standard 1-mark MCQ — just show the midpoint step and state the answer.

Mathematics — Standard (041) — CBSE Class 10 board question