Question

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Q.7

Answer 1

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Answer 2

Question:- Find the  point  on the axis which is  equidistant from (2, -5) and (-2,9)

Answer:

(-7,0)

Step-by-step explanation:

Let  the  point  on the  x- axis  be  P(x,0)

A = (2,-5)

B = (-2,9)

According  to question,

PA = PB

Making  Square of  both sides

PA² = PB²          ...............i)

Distance Between points P(x,0) and  A(2,-5):-

$PA=\sqrt{(x-2)^2+(0-(-5))^2}\\\;\\PA=\sqrt{(x-2)^2+5^2}\\\;\\PA=\sqrt{(x^2+4-4x+25)}\\\;\\PA=\sqrt{x^2-4x+29}\\\;\\(PA)^2=x^2-4x+29$

Distance Between points P(x,0) and  B(-2,9):-

$PB=\sqrt{(x+2)^2+(0-9)^2}\\\;\\PB=\sqrt{(x+2)^2+9^2}\\\;\\PB=\sqrt{(x^2+4+4x+81)}\\\;\\PB=\sqrt{x^2+4x+85}\\\;\\(PB)^2=x^2+4x+85$

Putting  these  values  in Eq. i)

PA² = PB²

$x^2-4x+29=x^2+4x+85\\\;\\\implies -4x+29=4x+85\\\;\\\implies 29-85=4x+4x\\\;\\\implies-56=8x\\\;\\\implies x=-7$

Hence,

Coordinates  of  Point  P = (-7,0)