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

Learn Exercise 7.1

We have to find a point on x-axis. Therefore, its y-coordinate will be 0. Let the point on x-axis be. Distance between (x,0) and (2,-5) = Distance between (x,0) and (-2,9)= By the given condition, these distances are equal in measure. x=-7 Therefore, the point is (− 7, 0).

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.

Let the point on x-axis be.

Distance between (x,0) and (2,-5) =$\sqrt{(x-2)^{2}+(0-(-5))^{2}}=&space;\sqrt{(x-2)^{2}+(5)^{2}}$

Distance between (x,0) and (-2,9)= $\sqrt{(x-(-2)^{2}+(0-(-9))^{2}}=&space;\sqrt{(x-2)^{2}+(9)^{2}}$

By the given condition, these distances are equal in measure.

$\sqrt{(x-2)^{2}+(5)^{2}}=&space;\sqrt{(x-2)^{2}+(9)^{2}}$

$(x-2)^{2}+25=(x+2)^{2}+81$

x=-7

Therefore, the point is (− 7, 0).

Professor with Masters degree (USA)

(0,2)

Tutor

Using midpoint formula we get (0,2) as answer so they mentioned point on x axis so it can be the 0

I have more than 12yrs experience in teaching side

Let the given points be P(2,-5), Q(-2,9) And the points required be R(a,0) The points is on the x-axis, y=0 And assuming x=a Hence the point is (a,0) As per question, Point R is equidistant from P & Q Hence PR = QR

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Balaji

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