Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.
Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.
The vertices of the given triangle are $A(4,4), B(3,5),$ and $C(-1,-1)$.
It is known that the slope $(m)$ of a non-vertical line passing through the points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2},\right.$ $y $$_{2}$ $)$ is given by $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, x_{2} \neq x_{1}$
$\therefore$ Slope of $AB \left( m _{1}\right)=\frac{5-4}{3-4}=-1$
Slope of $BC \left( m _{2}\right)=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$
Slope of $CA \left( m _{3}\right)=\frac{4+1}{4+1}=\frac{5}{5}=1$
It is observed that $m _{1} m _{3}=-1$
This shows that line segments $AB$ and $CA$ are perpendicular to each other i.e., the given triangle is right-angled at $A (4,4)$
Thus, the points $(4,4),(3,5),$ and $(-1,-1)$ are the vertices of a right-angled triangle.
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