Question

Show that $A(2,3),B(4,5)$ and $C(3,2)$ can be the vertices of a rectangle. Find the coordinates of the fourth vertex.

Answer 1

Given: $A=(2,3);B=(4,5);C=(3,2)$

Slope of $AB=\frac{5-3}{4-2}=1$

Slope of $BC=\frac{2-5}{3-4}=\frac{-3}{-1}=3$

Slope of $AC=\frac{2-3}{3-2}=-1$

Slope of $AB\times$ Slope of $AC=-1$

$\therefore AB\perp AC$

Let $D=(x,y)$

Slope of AC=Slope of BD

$\therefore \frac{3-2}{2-3}=\frac{5-y}{4-x}$

$\Rightarrow 5-y=x-4\Rightarrow x+y-9=0$

Also, Slope of $BD\times$ Slope of $DC=-1$

$\Rightarrow \frac{5-y}{4-x}\times \frac{y-2}{x-3}=-1$

$\Rightarrow (5-y)(y-2)=(3-x)(4-x)$

$\Rightarrow 5y-10-y^2+2y=12-3x-4x+x^2$

$\Rightarrow y^2-7y+10=-x^2+7x-12$

Puttig $y=9-x$

${(9-x)}^2-7(9-x)+10+x^2-7x+12=0$

$\Rightarrow 81+x^2-18x-63+7x+10+x^2-7x+10=0$

$\Rightarrow 2x^2-18x+40=0$

$x^2-9x+20=0\Rightarrow x=5$ or $x=4$

If $x=5,y=4$

If $x=4,y=5$

$(4,5)$ is already a vertex (B)

$\therefore D=(5,4)$

$\therefore$Four vertices are $A(2,3);B(4,5);C(3,2);D(5,4)$