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

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Solution

The given points are A(4,7), B(p,3) and C(7,3).

Since A,B and C are the vertices of a right angled triangle then, $(A B)^{2} + (B C)^{2} = (A C^{2})$

[By Pythagoras theorem]

$[(p - 4)^{2} + (3 - 7)^{2}] + [(7 - p)^{2} + (3 - 3)^{2}] = [(7 - 4)^{2} + (3 - 7)^{2}]$

$(p - 4)^{2} + (- 4)^{2} + (7 - p)^{2} = (3)^{2} + (- 4)^{2}$

$p^{2} + 16 - 8 p + 16 + 49 + p^{2} - 14 p = 9 + 16$

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

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

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

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

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

$p = 4 o r 7$

$p \neq 7$ (As B and C will coincide)

So, $p = 4$

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