Question

# If the point $$A(0,2)$$ is equidistant from the points $$B(3,p)$$ and $$C(p,5)$$, find $$\text p$$. Also, find the length of $$AB$$.

Solution

## Given,$$AB=AC$$$$(AB)^{2} = (AC)^{2}$$Distance between two points=$$\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$$$$(AB)^{2} = (AC)^{2} \implies$$$$(0-3)^{2}+(2-p)^{2} = (0-p)^{2} +(2-5)^{2}$$$$9+4+p^{2}-4p=p^{2}+9$$$$p= 1$$$$Distance = \sqrt {(0-3)^{2} + (2-1)^{2}}$$$$Distance = \sqrt {10}$$Mathematics

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