Question

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

{ALIGARH MUSLIM UNIVERSITY BOARD OF SECONDARY AND SENIOR SECONDARY EDUCATION, SAMPLE PAPER}

Answer 1

HELLO DEAR,

the points on X-axis must be CORDINATE of
y is 0

let the poin of x-axis be p(x,0)

given that :-A(2,-5) and B (-2,9). is equidistant from poin p


then


PA=PB

PB²=PA²

√[(x2 - x1)² + (y2 - y1)²]
distance OF AP

$\sqrt{( {2 - x)}^{2} + ( { - 5 - 0)}^{2} } \\ = > 4 - 4x + {x}^{2} + 25 \\ = > {x}^{2} - 4x + 29$
distance of PB

$\sqrt{( { - 2 - x)}^{2} + ( {9 - 0)}^{2} } \\ = > 4 + 4x + {x}^{2} + 81 \\ = > {x}^{2} + 4x + 85$
given that:-

AP=PB

${x }^{2} - 4x + 29 = {x}^{2} + 4x + 85 \\ = > - 8x = 56 \\ = > x = \frac{56}{8} \\ = > x = - 7$
then
P(-7,0)

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Step-by-step explanation:

Hey Mate......

The point is on the x axis. Hence we can assume the coordinate of the point as (h,0).

Now the point is equidistant from (2,-5) and (-2,9).

So

Simplifying this we get

Or 8h = -56 or h= -7

Hence the point is (-7,0) which is equidistant from (2,-5) and (-2,9).

hope it's helpful to you....

mark as brainliest......