In the diagram below, PQR is a triangle with vertices P(-5; -2), Q(-1; -6) and R(5; 4) - NSC Technical Mathematics - Question 1 - 2024 - Paper 2

N is the midpoint of RQ. The coordinates of R are (5, 4) and those of Q are (-1, -6). The midpoint formula is:

$N \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Thus,

$N = \left(\frac{5 + (-1)}{2}, \frac{4 + (-6)}{2}\right) = \left(\frac{4}{2}, \frac{-2}{2}\right) = (2, -1)$

Hence, the coordinates of N are (2, -1).

Since we found earlier that the gradient of PQ is -1, the equation of a line in point-slope form is:

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

Substituting R(5, 4) into this equation, we get:

$y - 4 = -1(x - 5)$

Expanding this,

$y - 4 = -x + 5$

So,

$y = -x + 9$

This is the equation of the line parallel to PQ passing through R.

To find the length of PQ, we use the distance formula:

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Using the coordinates P(-5, -2) and Q(-1, -6), we calculate:

$PQ = \sqrt{((-1) - (-5))^2 + ((-6) - (-2))^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2}$

For SN, we find the distance between S(0, 1) and N(2, -1):

$SN = \sqrt{(2 - 0)^2 + ((-1) - 1)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$

Now, since:

$2 \times SN = 2 \times (2 \sqrt{2}) = 4 \sqrt{2} = PQ$

Thus, we have shown that PQ = 2 SN.