Show that the triangle with vertices at \( P(-5,6), Q(2,3) \) and \( R(5,10) \) is a right-angled triangle.

To show that the triangle \( PQR \) is a right-angled triangle, we will check the slopes of the sides formed by the vertices and verify if they are perpendicular, indicating a right angle. 1. Calculate the slopes: - Slope of \( PQ \): \[ m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 6}{2 - (-5)} = \frac{-3}{7} \] - Slope of \( QR \): \[ m_{QR} = \frac{10 - 3}{5 - 2} = \frac{7}{3} \] - Slope of \( RP \): \[ m_{RP} = \frac{6 - 10}{-5 - 5} = \frac{-4}{-10} = \frac{2}{5} \] 2. Check if any pairs of slopes are negative reciprocals: - For \( PQ \) and \( QR \): \[ m_{PQ} \times m_{QR} = \left( -\frac{3}{7} \right) \times \left( \frac{7}{3} \right) = -1 \quad \text{(perpendicular)} \] - For \( QR \) and \( RP \): \[ m_{QR} \times m_{RP} = \left( \frac{7}{3} \right) \times \left( \frac{2}{5} \right) = \frac{14}{15} \quad \text{(not perpendicular)} \] - For \( RP \) and \( PQ \): \[ m_{RP} \times m_{PQ} = \left( \frac{2}{5} \right) \times \left( -\frac{3}{7} \right) = -\frac{6}{35} \quad \text{(not perpendicular)} \] Since \( PQ \) and \( QR \) are perpendicular, triangle \( PQR \) is a right-angled triangle with the right angle at vertex \( Q \). Thus, we have shown that triangle \( PQR \) is indeed a right-angled triangle.