Question

19. Find a point on the x-axis which is equidistant from the points (2,-5) and (-2,9).​

Answer 1

Answer:

The point on x-axis is (-7,0)

Step-by-step explanation:

A point on x-axis means (x,0)

Let P(x,0) A(2,-5) B(-2,9)

According to question

PA=PB

Using distance formula

PA=$\sqrt{ (x_{2}-x_{1})^2+(y_{2}-y_{1})^{2}$        

PA=$\sqrt{(2-x)^{2}+(-5-0)^{2}$

=$\sqrt{2^{2}+x^2-4x+25$

=$\sqrt{4+x^2-4x+25$

=$\sqrt{x^2-4x+29}........Eq-1$

PB=$\sqrt{ (x_{2}-x_{1})^2+(y_{2}-y_{1})^{2}$

=$\sqrt{(-2-x)^2+(9-0)^2$

=$\sqrt{ (-2)^2+x^2-2(-2)(x)+81$

=$\sqrt{4+x^2+4x+81$

=$\sqrt{x^2+4x+85}........Eq-2$

PA=PB

$\sqrt{x^2-4x+29}$=$\sqrt{x^2+4x+85}$

Simplifying it

${x^2-4x+29}$=${x^2+4x+85}$

$x^2-x^{2} -4x-4x+29-85=0$

$-8x-56=0\\-8x=56\\x=56/-8\\x=-7$

∴The point on x-axis is (-7,0)

Thank you