Question

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

Answer 1

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 (x1, y1) and (x2, y2) is given by.

$m = \frac {y_2-y_1}{x_2-x_1}, \ x_2≠x_1$

∴ Slope of AB $(m_1) = \frac {5-4}{3-4} = -1$

Slope of BC $(m_2) = \frac {-1-5}{-1-3}= \frac {-6}{-4} = \frac 32$

Slope of CA $(m_3) =\frac { 4+1 }{ 4+1 }= \frac 55 = 1$

It is observed that $m_1m_3 = -1$

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.