Question

Find the value of $$k$$ such that the points $$\mathrm{P}(k, 1), \mathrm{Q}(2, -5)$$ and $$\mathrm{R}(k -2, -3)$$ are collinear.

Answer 1

Since the points $$P$$, $$Q$$ and $$R$$ are collinear, the slope of $$PQ$$ must be equal to the slope of $$QR$$.

Slope of $$PQ$$: $$m_{PQ} = \frac{-5 - 1}{2 - k}$$

$$= \frac{-6}{2 - k}$$

Slope of $$QR$$: $$m_{QR} = \frac{-3 - (-5)}{(k - 2) - 2}$$

$$= \frac{-3 + 5}{k - 4}$$

$$= \frac{2}{k - 4}$$

Equating the two slopes: $$\frac{-6}{2 - k} = \frac{2}{k - 4}$$

Dividing both numerators by $$2$$: $$\frac{-3}{2 - k} = \frac{1}{k - 4}$$

Cross-multiplying gives: $$-3(k - 4) = 1(2 - k)$$

$$-3k + 12 = 2 - k$$

$$-3k + k = 2 - 12$$

$$-2k = -10$$

$$k = 5$$

Thus, the value of $$k$$ is $$5$$.