The fourth vertex $ \mathrm{D} $ of a parallelogram $ \mathrm{ABCD} $ whose three vertices are $ \mathrm{A}(-2,3), \mathrm{B}(6,7) $ and $ \mathrm{C}(8,3) $ is
(A) $ (0,1) $
(B) $ (0,-1) $
(C) $ (-1,0) $
(D) $ (1,0) $

Given:

Three vertices of a parallelogram \( \mathrm{ABCD} \) are \( \mathrm{A}(-2,3), \mathrm{B}(6,7) \) and \( \mathrm{C}(8,3) \)

To do:

We have to find the fourth vertex \( \mathrm{D} \).

Solution:

Let the fourth vertex of the parallelogram be $D(x_4 ,y_4)$ and the 

We know that,

The diagonals of a parallelogram bisect each other.

Let the point of intersection of the diagonals be $O(x, y)$

The mid-point of the line segment joining the points $A (-2, 3)$ and $C(8, 3)$ is,

Using mid-point formula, we have

$O( x,\ y)=( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$

$=( \frac{-2+8}{2}, \frac{3+3}{2})$

$=( \frac{6}{2}, \frac{6}{2})$

$=( 3, 3)$

$O(x,y)$ is also the mid-point of the line segment joining the points $B(6, 7)$ and $D(x_4, y_4)$

This implies,

$O(3, 3)=( \frac{6+x_4}{2}, \frac{7+y_4}{2})$

$\Rightarrow 3=\frac{6+x_4}{2}, 3= \frac{7+y_4}{2}$

$6+x_4=3(2), 7+y_4=3(2)$

$x_4=6-6, y_4=6-7$

$x_4=0, y_4=-1$

The fourth vertex \( \mathrm{D} \) is $(0, -1)$.

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Updated on: 10-Oct-2022

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