PQR is a triangle with vertices P(0, -4), Q(-6, 2), and R(10, 6) - Scottish Highers Maths - Question 1 - 2016

The coordinates of point P are (0, -4) and M are (2, 4). First, find the slope (m) of line PM:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-4)}{2 - 0} = \frac{8}{2} = 4.$

Using the point-slope form of a line equation, which is:

$y - y_1 = m(x - x_1),$

the equation of line PM is:

$y - (-4) = 4(x - 0) \Rightarrow y + 4 = 4x \Rightarrow y = 4x - 4.$

First, we need to find the slope of line PR. The coordinates of P are (0, -4) and R are (10, 6):

$m_{PR} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-4)}{10 - 0} = \frac{10}{10} = 1.$

The slope of line L, which is perpendicular to PR, is the negative reciprocal:

$m_L = -\frac{1}{m_{PR}} = -1.$

Using M(2, 4) to find the equation of line L:

$y - 4 = -1(x - 2) \Rightarrow y - 4 = -x + 2 \Rightarrow y = -x + 6.$

First, we need to find the midpoint of PR:

$Midpoint_{PR} = \left( \frac{x_P + x_R}{2}, \frac{y_P + y_R}{2} \right) = \left( \frac{0 + 10}{2}, \frac{-4 + 6}{2} \right) = \left( 5, 1 \right).$

Now, we check if this point satisfies the equation of line L:

Substituting x = 5 into line L's equation:

$y = -5 + 6 = 1.$

Since the y-coordinate matches, line L passes through the midpoint of PR.