The points $A (4, 7), B (p, 3)$ and $C (7, 3)$ are the vertices of a right triangle, right-angled at $B$ , Find the values of $P$ .

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Solution

Given $A (4, 7), B (p, 3)$ and $C (7, 3)$

Distance Formula $= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

$A B = \sqrt{(p - 4)^{2} + (3 - 7)^{2}} = \sqrt{(p - 4)^{2} + 16}$

$B C = \sqrt{(7 - p)^{2} + (3 - 3)^{2}} = \sqrt{(7 - p)^{2}}$

$A C = \sqrt{(7 - 4)^{2} + (3 - 7)} = \sqrt{(3)^{2} + (4)^{2}} = 5$

Triangle $A B C$ is right angle at $B$ , hence

$(A C)^{2} = (A B)^{2} + (B C)^{2}$

$∴ {(\sqrt{(p - 4)^{2} + 16})}^{2} + {(\sqrt{(7 - p)^{2}})}^{2} = 25$

$\Rightarrow p^{2} - 8 p + 16 + 16 + 49 - 14 p + p^{2} = 25$

$\Rightarrow 2 p^{2} - 22 p - 81 = 25$

$\Rightarrow 2 p^{2} - 22 p + 56 = 0$

$\Rightarrow p^{2} - 11 p + 28 = 0$

$\Rightarrow p^{2} - 7 p - 4 p + 28 = 0$

$\Rightarrow p (p - 7) - 4 (p - 7) = 0$

Or $(p - 7) (p - 4) = 0$

If $p - 7 = 0$ then $p = 7$ , which is not possible as $B$ and $C$ will be same.

If $p - 4 = 0$ then $p = 4$

Hence, $p = 4$ .

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