Without using the Pythagoras theorem,show that the points $(4,4), (3,5),$ and $(-1,-1)$ are vertices of a right-angled triangle.
(N/A) Let the vertices of the triangle be $A(4,4), B(3,5),$ and $C(-1,-1)$.
The slope $(m)$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Slope of $AB$ $(m_1) = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1$.
Slope of $BC$ $(m_2) = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}$.
Slope of $CA$ $(m_3) = \frac{4 - (-1)}{4 - (-1)} = \frac{5}{5} = 1$.
Since the product of the slopes of $AB$ and $CA$ is $m_1 \times m_3 = (-1) \times (1) = -1$,the lines $AB$ and $CA$ are perpendicular.
Therefore,the triangle is a right-angled triangle at vertex $A(4,4)$.
The slope $(m)$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Slope of $AB$ $(m_1) = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1$.
Slope of $BC$ $(m_2) = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}$.
Slope of $CA$ $(m_3) = \frac{4 - (-1)}{4 - (-1)} = \frac{5}{5} = 1$.
Since the product of the slopes of $AB$ and $CA$ is $m_1 \times m_3 = (-1) \times (1) = -1$,the lines $AB$ and $CA$ are perpendicular.
Therefore,the triangle is a right-angled triangle at vertex $A(4,4)$.
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