Using slopes show that the points A


Using slopes show that the points A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) taken in order, are the vertices of a rectangle.


A rectangle has all sides perpendicular to each other, so the product of slope of every adjacent line is equal to -1.

Given point in order are $A(-4,-1), B(-2,-4), C(4,0)$ and $D(2,3)$

slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$

Slope of $A B=\left(\frac{-4+1}{-2+4}\right)=\frac{-3}{2}$

Slope of $B C=\left(\frac{0+4}{4+2}\right)=\frac{4}{6}=\frac{2}{3}$

The slope of $C D=\left(\frac{3-0}{2-4}\right)=\frac{3}{-2}$

Slope of $D A=\left(\frac{3+1}{2+4}\right)=\frac{4}{6}=\frac{2}{3}$

⇒slopeofAB × slopeofBC

$\Rightarrow \frac{-3}{2} \times \frac{2}{3}=-1$

Hence AB is perpendicular to BC

Slope of BC × slope of CD

$\frac{2}{3} \times \frac{3}{-2}=-1$

Hence BC is perpendicular to CD

Slope of CD × slope of DA

$\Rightarrow \frac{3}{-2} \times \frac{2}{3}=-1$

Hence $C D$ is perpendicular to $D A$

Slope of $D A \times$ slope of $A B$

$\Rightarrow \frac{2}{3} \times \frac{-3}{2}=-1$

Hence DA is perpendicular to AB.

All angles are $90^{\circ}$.

So this is a rectangle $A B C D$.


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