The point on the X- axis which is equidistant | vedclass

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Question

(A) Let the point on the $X-$ axis be $P(x, 0)$.Since $P$ is equidistant from $A(-1, 2)$ and $B(5, 4)$,we have $PA = PB$,which implies $PA^2 = PB^2$.U

Vedclass · Accepted Answer

$(3, 0)$

Vedclass · Answer

(A) Let the point on the $X-$ axis be $P(x, 0)$.Since $P$ is equidistant from $A(-1, 2)$ and $B(5, 4)$,we have $PA = PB$,which implies $PA^2 = PB^2$.Using the distance formula,we get:$[x - (-1)]^2 + (0 - 2)^2 = (x - 5)^2 + (0 - 4)^2$$(x + 1)^2 + 4 = (x - 5)^2 + 16$$x^2 + 2x + 1 + 4 = x^2 - 10x + 25 + 16$$2x + 5 = -10x + 41$$12x = 36$$x = 3$Therefore,the required point on the $X-$ axis is $(3, 0)$.

The point on the $X-$ axis which is equidistant from $A (-1, 2)$ and $B(5, 4)$ is $\ldots \ldots \ldots \ldots$

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