**Question:**

Without using Pythagora’s theorem, show that the points A(1, 2), B(4, 5) and C(6, 3) are the vertices of a right-angled triangle.

**Solution:**

The ΔABC is made up of three lines, AB,BC and CA

For a right angle triangle, two lines must be at $90^{\circ}$ so they are perpendicular to each other.

Checking for lines AB and BC

For two lines to be perpendicular, their product of slope must be equal to -1.

Given points A(1, 2), B(4, 5) and C(6, 3)

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

Slope of $A B=\left(\frac{5-2}{4-1}\right)=\frac{3}{3}=1$

Slope of $B C=\left(\frac{3-5}{6-4}\right)=\frac{-2}{2}=-1$

Slope of $\mathrm{CA}=\left(\frac{3-2}{6-1}\right)=\frac{1}{5}=0.2$

Checking slopes of line AB and BC

1×-1 = -1

So $A B$ is Perpendicular to $B C$.

So it is a right angle triangle.