Question

Three vertices of parallelogram ABCD taken in order are A(3,6), B(5, 10) and C(3,2).
(i) the coordinate of the fourth vertex D
(ii) length of diagonal BD
(iii) equation of the side AD of the parallelogram ABCD

Answer 1

Let the vertex $D$ have coordinates $(x,y)$

Midpoint of diagonal $BD$ & diagonal $AC$ is the same

$\therefore$ By Midpoint formula

$\frac{x+5}{2}=\frac{3+3}{2}-$(Equating $x$-coordinate)

$\frac{y+10}{2}=\frac{6+2}{2}-$(Equating $y$-coordinate)

$\therefore x=1y=2$

$l\left(BD\right)=\sqrt{(x_B-x_D{)}^2+(y_B-y_D{)}^2}$

$=\sqrt{(5-1{)}^2+(10-(-2){)}^2}=\sqrt{16+144}=\sqrt{160}$

$=4\sqrt{10}$

Equation od side $AD$

by double coordinate form of equation of line,

$y-y_A=\left(\frac{y_D-y_A}{x_D-x_A}\right)(x-x_A)$

$y-6=\frac{-2-6}{1-3}(x-3)$

$\therefore y-6=4(x-3)$

$\therefore 4x-y=6$