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

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(A) We need to find a point on the $x$-axis. Therefore,its $y$-coordinate must be $0$.Let the point on the $x$-axis be $P(x, 0)$.The distance between

Vedclass · Accepted Answer

$(-7, 0)$

Vedclass · Answer

(A) We need to find a point on the $x$-axis. Therefore,its $y$-coordinate must be $0$.Let the point on the $x$-axis be $P(x, 0)$.The distance between $P(x, 0)$ and $A(2, -5)$ is given by the distance formula:$PA = \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - 2)^2 + 25}$.The distance between $P(x, 0)$ and $B(-2, 9)$ is given by:$PB = \sqrt{(x - (-2))^2 + (0 - 9)^2} = \sqrt{(x + 2)^2 + 81}$.Since the point is equidistant from $A$ and $B$,we have $PA = PB$,which implies $PA^2 = PB^2$:$(x - 2)^2 + 25 = (x + 2)^2 + 81$Expanding the squares:$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$Subtracting $x^2 + 4$ from both sides:$-4x + 25 = 4x + 81$Rearranging the terms:$-8x = 81 - 25$$-8x = 56$$x = -7$Thus,the required point is $(-7, 0)$.

Find the point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9).$

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