The point on x-axis which is equidistant from the points $(5,-3)$ and $(4,2)$ is :\

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Professor PC Thomas Classes · Answer

(b) : Let point on $x$-axis which is equidistant from points $(5,-3)$ and $(4,2)$ be $(x, 0)$.\ $\Rightarrow \sqrt{(5-x)^{2}+(-3-0)^{2}}=\sqrt{(4-x)^{2}+(2-0)^{2}}$\ $\Rightarrow(5-x)^{2}+9=(4-x)^{2}+4$\ $\Rightarrow 25+x^{2}-10 x+9=16+x^{2}-8 x+4$\ $\Rightarrow 34-10 x=20-8 x \quad \Rightarrow \quad 14=2 x$\ $\Rightarrow \quad x=7$, So, required point is $(7,0)$.