Question

find the point on x-axis which is equidistant form the points ( 2,-2) and(-4,2).


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Answer 1

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

Two points are given :A(2,-2) ,B(-4,2)

To Find :Point on x -axis which is equidistant from the points A and B

Let us assume that the point be P(x,0) which is equidistant from the given points A and B

=>Let x be 'a'

Remember:On x-axis y's value is zero.Similarly on Y-axis X's value is zero.

If Point P is equidistant from two two points A and B then :

PA =PB

Here, distance formula will used to find out the distance between the two given Points.

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

A(2,-2) ,P(a,0) and B(-4,2)

$\sf \: PA = \sqrt{ {(a - 2)}^{2} + {(0 + 2)}^{2} }$

Formula:(a-b)²=a²+b²-2ab

$\sf \: PA= \sqrt{ {a}^{2} + 4 - 4a + 4}$

$\sf \: PA = \sqrt{ {a}^{2} - 4a + 8}$

$\sf \: PB= \sqrt{ {[ - (4 + a)]}^{2} + {(2 - 0)}^{2} }$

$\sf \: PB= \sqrt{ {(4 + a)}^{2} + {2}^{2} } = \sqrt{ {4}^{2} + {a}^{2} + 8a + 4 }$

$\sf \: PB = \sqrt{16 + {a}^{2} + 8a + 4} = \sqrt{ {a}^{2} + 8a + 20}$

Now,PA=PB

squaring both sides

(PA)²=(PB)²

When we square the roots will removed:

=>a²-4a+8=a²+8a+20

a² lies on both side so,it gets cancelled

=>-4a+8=8a+20

=>-4a-8a=20-8

=>-12a=12

a=-1

Our point is P(a,0)

So,∴ Our point is P(-1,0)

∴The point on x-axis which is equidistant from the point A(2,-2) and B(-4,2) is P(-1,0)