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The points A$(3, 6)$, B$(k, 2)$ and C$(6, 2)$ are the vertices of a right triangle ABC right-angled at B. Find the value of $k$.
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AC$^2 = (6 – 3)^2 + (2 – 6)^2 = 25$
AB$^2 = (k – 3)^2 + (2 – 6)^2 = k^2 – 6k + 25$
BC$^2 = (6 – k)^2 + (2 – 2)^2 = k^2 – 12k + 36$
Since $\triangle$ ABC is right angled at B.
$\therefore AB^2 + BC^2 = AC^2$
$\Rightarrow k^2 - 9k + 18 = 0$
$\Rightarrow (k – 3)(k – 6) = 0$
$\Rightarrow k = 3, 6$
AB$^2 = (k – 3)^2 + (2 – 6)^2 = k^2 – 6k + 25$
BC$^2 = (6 – k)^2 + (2 – 2)^2 = k^2 – 12k + 36$
Since $\triangle$ ABC is right angled at B.
$\therefore AB^2 + BC^2 = AC^2$
$\Rightarrow k^2 - 9k + 18 = 0$
$\Rightarrow (k – 3)(k – 6) = 0$
$\Rightarrow k = 3, 6$