The locus of the third vertex of a right-angl | vedclass

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(A) Let the third vertex be $P(x, y)$.Since the triangle is right-angled at $P$,the angle $\angle P = 90^{\circ}$.This implies that the point $P$ lies

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$x^2+y^2-5x-7y+14=0$

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(A) Let the third vertex be $P(x, y)$.Since the triangle is right-angled at $P$,the angle $\angle P = 90^{\circ}$.This implies that the point $P$ lies on a circle where the hypotenuse joining $(1, 2)$ and $(4, 5)$ is the diameter.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting the given points $(1, 2)$ and $(4, 5)$:$(x-1)(x-4) + (y-2)(y-5) = 0$$x^2 - 5x + 4 + y^2 - 7y + 10 = 0$$x^2 + y^2 - 5x - 7y + 14 = 0$.Note: The vertex $P$ cannot be $(1, 2)$ or $(4, 5)$ as it would not form a triangle.

The locus of the third vertex of a right-angled triangle,the ends of whose hypotenuse are $(1,2)$ and $(4,5)$,is

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