If $A (9, - 9), C (1, - 3)$ are the end points of hypotenuse of a right angled isosceles triangle, then the third vertex is

(8, - 2)

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(- 8, 2)

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(8, - 8)

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(0, 0)

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Solution

The correct option is A $(8, - 2)$
$s l o p e o f A B \times s l o p e o f B C = - 1$
$\Rightarrow \frac{- 9 - k}{0 - n} \times \frac{- 3 - k}{1 - n} = - 1$
$\Rightarrow (9 + k) (3 + k) = (n - 1) (9 - n) . . (i)$
$A B^{2} + B C^{2} = A C^{2}$
$\Rightarrow B C^{2} + B C^{2} = 64 + 36$
$\Rightarrow 2 B C^{2} = 100$
$\Rightarrow B C^{2} = 50$
$\Rightarrow (n - 1)^{2} + (k + 3)^{2} = 50 . . . (i i)$
clearly $(8, - 2)$ satisfy both the equation
$∴ B (8, - 2)$

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If A(9,−9),C(1,−3) are the end points of hypotenuse of a right angled isosceles triangle, then the third vertex is

The correct option is A (8,−2)slopeofAB×slopeofBC=−1⇒−9−k0−n×−3−k1−n=−1⇒(9+k)(3+k)=(n−1)(9−n)..(i)AB2+BC2=AC2⇒BC2+BC2=64+36⇒2BC2=100⇒BC2=50⇒(n−1)2+(k+3)2=50...(ii)clearly (8,−2) satisfy both the equation∴B(8,−2)

If $A (9, - 9), C (1, - 3)$ are the end points of hypotenuse of a right angled isosceles triangle, then the third vertex is