Question

The points $A(1,-2),B(2,3),C(k,2)$ and $D(-4,-3)$ are the vertices of a parallelogram. Find the value of $k$.

• A. $-2$
• B. $2$
• C. $3$
• D. $-3$

Answer 1

The correct option is D $-3$

We know that diagonals of a parallelogram bisects each other.

So, the co-ordinates A$(1,-2)$ B$(2,3)$, C$(k,2)$ and D$(-4,-3)$

$\therefore$ Mid-point of AC$=$Mid point of BD

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(\frac{1+k}{2},\frac{-2+2}{2})=(\frac{2+(-4)}{2},\frac{3+(-3)}{2})$

$(\frac{1+k}{2},0)=(\frac{-2}{2},\frac{0}{2})$

$\frac{1+k}{2}=\frac{-2}{2}$

$2(1+k)=-4$

$2+2k=-4$

$2k=-4-2$

$2k=-6$

$k=\frac{-6}{2}$

$\therefore k=-3$

$\therefore$ So, the value of $K=-3$.