Find the point on the X-axis which is equidis | vedclass

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(C) Let the point on the $X$-axis be $P(x, 0)$.Since $P$ is equidistant from $A(5, 4)$ and $B(-2, 3)$,we have $PA = PB$.Using the distance formula,$PA

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$(2, 0)$

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(C) Let the point on the $X$-axis be $P(x, 0)$.Since $P$ is equidistant from $A(5, 4)$ and $B(-2, 3)$,we have $PA = PB$.Using the distance formula,$PA^2 = PB^2$.$(x - 5)^2 + (0 - 4)^2 = (x - (-2))^2 + (0 - 3)^2$$(x - 5)^2 + 16 = (x + 2)^2 + 9$$x^2 - 10x + 25 + 16 = x^2 + 4x + 4 + 9$$-10x + 41 = 4x + 13$$41 - 13 = 4x + 10x$$28 = 14x$$x = 2$.Therefore,the point is $(2, 0)$.

Find the point on the $X$-axis which is equidistant from $(5, 4)$ and $(-2, 3)$.

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