Question

. Find the point on the X axis which is equidistant from A(-3,4) and B(1,-4) .

Solution: Let P(x, ---) be the point on X axis which is equidistant from points A and B.

∴ d(P,A)2

= d(P,B)2

x

2 + 6x + 9 = x2

– 2x +1 ______[ using -------- formula]

----- = -8

x = ------

∴ coordinates of the point on X axis which is equidistant from A and B are P(---, ----).​

Answer 1

Step-by-step explanation:

please give me thanks and mark me brilient

Answer 2

Given

Two points A(-3,4) B(1,-4)

To Find

Point which is equidistant from Given Point A and B

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Let the required point be P(x,0) which is equidistant from the two given points A(-3,4) and B(1,-4)

let x be a ,we takeP( x,0 )because on x axis y is zero

•A(-3,4)___________•P(a,0)___________•B(1,-4)

we will use distance formula for finding distance between any two points

$\sf \boxed{D= \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2 }- y_{1})}^{2}}}$

If P is equidistant from point A and B then AP=PB

AP=√(a+3)²+(0-4)²

AP=√a²+9+6a+16

AP=√a²+6a+25

PB=√(1-a)²+(-4-0)²

PB=√1+a²-2a+16

PB=√a²-2a+17

Now AP=PB

√a²+6a+25=√a²-2a+17

squaring both sides

=>a²+6a+25=a²-2a+17

a² gets cancelled as lie on both side [base same]

=>6a+2a+25-17=0

=>8a+8=0

=>8a=-8

a=-8/8=-1

a=-1

∴P(-1,0) is the point on x axis which is equidistant from A(-3,4) and B(1,-4)