Given two fixed points A(-2, 1) and B(3, 0),f | vedclass

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(B) Let the coordinates of point $P$ be $(h, k)$.Given points are $A(-2, 1)$ and $B(3, 0)$.The slope of $AP$ is $m_1 = \frac{k-1}{h+2}$.The slope of $

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$x^2+y^2-x-y-6=0$

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(B) Let the coordinates of point $P$ be $(h, k)$.Given points are $A(-2, 1)$ and $B(3, 0)$.The slope of $AP$ is $m_1 = \frac{k-1}{h+2}$.The slope of $BP$ is $m_2 = \frac{k-0}{h-3} = \frac{k}{h-3}$.Since $\angle APB = 90^{\circ}$,the product of the slopes must be $-1$,i.e.,$m_1 m_2 = -1$.$\frac{k-1}{h+2} 	imes \frac{k}{h-3} = -1$.$\frac{k^2-k}{h^2-h-6} = -1$.$k^2-k = -(h^2-h-6)$.$h^2+k^2-h-k-6 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus of point $P$ is $x^2+y^2-x-y-6=0$.

Given two fixed points $A(-2, 1)$ and $B(3, 0)$,find the locus of a point $P$ which moves such that the angle $\angle APB$ is always a right angle.

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