Question

Three vertices of a parallelogram $\text{ABCD}$ taken in order are $\text{A}(3,6)$, $\text{B}(5,10)$ and $\text{C}(3,2)$ find :
(i) the coordinates of the fourth vertex $\text{D}$.
(ii) length of diagonal $\text{BD}$.

Answer 1

Let the coordinate of $D$ be $(x,y)$. In a parallelogram, mid point of diagonal $AC$ coincides with the mid-point of diagonal $BD$.
Mid point of $AC=(\frac{3+3}{2},\frac{6+2}{2})=(3,4)$

Mid point of $BD=(\frac{x+5}{2},\frac{y+10}{2})$
Equating,
(i) $3=\frac{x+5}{2}$ and $4=\frac{y+10}{2}$
$⟹x=1$ and $y=-2$
Coordinates of $D(1,-2)$

(ii) $BD=\sqrt{{(5-1)}^2+{(10+2)}^2}$
$=\sqrt{16+144}$
$=\sqrt{160}$
$=4\sqrt{10}$ units.