1. The vertices of the triangle \( \triangle P Q R \) are \( P(5,10), Q(2,1) \) and \( R(11,-2) \). a. Find the lengths of \( P Q \) and \( Q R \). b. Show that \( P Q \) is perpendicular to \( Q R \). c. Find the midpoint of \( P R \). Label the midpoint \( M \). d. Find the equation of the line through the points \( Q \) and \( M \).

To find the lengths of \( PQ \) and \( QR \): a. The length of \( PQ \) can be calculated using the distance formula: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 - 5)^2 + (1 - 10)^2} = \sqrt{(-3)^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}. \] For \( QR \): \[ QR = \sqrt{(11 - 2)^2 + (-2 - 1)^2} = \sqrt{(9)^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}. \] b. To show that \( PQ \) is perpendicular to \( QR \), calculate the slopes. The slope \( m_{PQ} \) is given by: \[ m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 10}{2 - 5} = \frac{-9}{-3} = 3. \] The slope \( m_{QR} \): \[ m_{QR} = \frac{-2 - 1}{11 - 2} = \frac{-3}{9} = -\frac{1}{3}. \] Since \( m_{PQ} \cdot m_{QR} = 3 \cdot \left(-\frac{1}{3}\right) = -1 \), \( PQ \) is indeed perpendicular to \( QR \). c. The midpoint \( M \) of \( PR \) is given by the midpoint formula: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{5 + 11}{2}, \frac{10 + (-2)}{2} \right) = \left( \frac{16}{2}, \frac{8}{2} \right) = (8, 4). \] d. Now, we find the equation of the line through \( Q(2,1) \) and \( M(8,4) \). The slope \( m_{QM} \) is: \[ m_{QM} = \frac{4 - 1}{8 - 2} = \frac{3}{6} = \frac{1}{2}. \] Using the point-slope form, the equation of the line through point \( Q \): \[ y - y_1 = m(x - x_1) \implies y - 1 = \frac{1}{2}(x - 2). \] Simplifying gives: \[ y - 1 = \frac{1}{2}x - 1 \implies y = \frac{1}{2}x. \]