Question

Without using the Pythagoras theorem, show that the points $(4,4),(3,5)\text{and}(-1,-1)$ are the vertices of a right-angled triangle.

Answer 1

Step $1:$ Simplification of given data
Let the $3$ points of triangle be $A(4,4),B(3,5),C(-1,-1)$
We know that slope of a line through the points $(x_1,y_1)\&(x_2,y_2)$ is
$m=\frac{y_2-y_1}{x_2-x_1}$
Slope of $AB$
Here, $x_1=4,y_1=4,x_2=3,y_2=5$
Slope of $AB=\frac{5-4}{3-4}=\frac{1}{-1}=-1$

Slope of $BC$
Here, $x_1=3,y_1=5,x_2=-1,y_2=-1$
Slope of $BC=\frac{-1-\left(5\right)}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$
Slope of $AC$
Here, $x_1=4,y_1=4$
$x_2=-1,y_2=-1$
Slope of $AC=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1$

Step $2:$ Solve for proving
If product of slopes of two lines is $-1$, it means that the lines are perpendicular and it is a right-angled triangle.
Now, Slope of $AB\times$ Slope of $AC$
$=(-1)\times \left(1\right)$
$=-1$
i.e., Lines $AB\&AC$ are perpendicular
Hence,it is right-angled triangle.