Question

If two vertices of a parallelogram are $(3,2),(-1,0)$ and its diagonals meet at $(2,-5)$, find the other two vertices of the parallelogram.

Answer 1

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ is

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

O is the midpoint of $AC$ and $BD$

$\therefore (2,-5)=$ mid point of $AC$

$(2,-5)=(\frac{x_1+3}{2},\frac{y_1+2}{2})$

$\frac{x_1+3}{2}=2$

$\Rightarrow x_1+3=4$

$\therefore x_1=1$

Also, $\frac{y_1+2}{2}=-5$

$y_1+2=-10$

$y_1=-12$

$\therefore c=(1,-12)$

$(2,-5)=$ Mid point of BD

$(2,-5)=(\frac{-1+x_2}{2},\frac{0+y_2}{2})$

$\Rightarrow \frac{-1+x_2}{2}=2$

$\Rightarrow -1+x_2=4$

$\therefore x_2=5$

$\Rightarrow \frac{0+y_2}{2}=-5$

$\therefore y_2=-10$

$\therefore D=(5,-10)$