gradient of a line 2

The point $A(4,5)$, $B(10,5)$ and $C(8,k)$ are such that angle $ACB$ is a right angle. Determine the two possible values of $k$

solution - press button to display

If $ACB$ is a right angle, then the line segments $AC$ and $BC$ are perpendicular, so $$ m_{AC} \times m_{BC} = -1 $$

The gradients are: $$m_{AC} = \frac{k-5}{8-4} = \frac{1}{4}(k-5),\; m_{BC} = \frac{k-5}{8-10} = -\frac{1}{2}(k-5)$$ and so $m_{AC}\times m_{BC}$ is $$ m_{AC}\times m_{BC} = -\frac{1}{8}(k-5)^2 $$ It follows that $-1 = -\frac{1}{8}(k-5)^2$, this gives $(k-5)^2 = 8$ and so $k = 5\pm 2\sqrt{2}$