Question

If three vertices of $\square$ $ABCDareA\left(1,2\right),B\left(2,1\right),C\left(3,-4\right)$, then co-ordinates of $D$ are

• A. $(2,-3)$
• B. $(4,-6)$
• C. $(0,1)$
• D. $(4,-2)$

Answer 1

The correct option is A $(2,-3)$

Given points are $A(1,2),B(2,1),C(3,-4)$ are the vertices of a parallelogram.

So, mid point of AC$=$mid point of BD

$A(\underset{x}{1},\underset{y}{2})$, $C(\underset{x_2}{3},\underset{y_2}{-4})=B(\underset{x}{2},\underset{y}{1})D(\underset{x_2}{x},\underset{y_2}{y})$

$[\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]=[\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]$

$\Rightarrow [\frac{1+3}{2},\frac{2-4}{2}]=[\frac{x+2}{2},\frac{1+y}{2}]$

$\frac{x+2}{2}=\frac{4}{2}$ $\frac{1+y}{2}=\frac{-2}{2}$

$x+2=4$ $1+y=-2$

$x=4-2$ $y=-2-1$

$x=2$ $y=-3$

$\therefore D(x,y)=(2,-3)$

Hence, correct option is $A(2,-3)$.