Question

6
(1) Find the point on X axis which is equidistant from the points M(-5,-2) and
N(3,2).​

Answer 1

Given

We have given two points M( -5,-2) and N( 3,2)

To find

we have to find the point on x - axis which is equidistant from the points M and N

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

since ,we don't know the points let us assume the point on x axis be ( x,0)

Note : On x-axis the y will be zero and on Y-axis x will be zero.

let x be 'a'

so , our new point be P (a,0)

According to the question:

If the point P is equidistant from M and N

then their distance between the points PM and PN must be equal.

Now, by using distance formula we will find distance between the two points.

Distance Between P and M

Distance formula

D=√(x₂- x₁)²+(y₂- y₁)²

M( -5,-2) and P(a,0)

x₁= -5 ;x₂ = a ; y₁= -2 & y₂ = 0

PM= √ (a+5)²+(0+2)²

identity :(a+b)²= a²+b²+2ab

PM= √ a²+5²+2(a)(5)+2²

PM= √a²+25+10a+4=√ a²+10a+29

Distance between P and N

P(a,0) N ( 3,2)

PN=√ (3-a)²+(2-0)²

PN= √ 9+a²-6a+4= √ a²-6a+13

Now , comparing both PM and PN

√ a²+10a+29=√a²-6a+13

Squaring both sides

PM²= PN²

(√a²+10a+29)²= (√a²-6a+13)²

a²+10a+29=a²-6a+13

a² gets cancelled as base same

=> 10a +29= -6a+13

=> 10a+6a=13-29

=> 16a = -16

=>a = -1

Thus,the point is (a,0) = (-1,0)

Therefore, (-1,0) is the point on the x-axis which is equidistant from the points M and N

Answer 2

Answer❣

Given

We have given two points M( -5,-2) and N( 3,2)

To find

we have to find the point on x - axis which is equidistant from the points M and N

Solution

since ,we don't know the points let us assume the point on x axis be ( x,0)

Note : On x-axis the y will be zero and on Y-axis x will be zero.

let x be 'a'

so , our new point be P (a,0)

According to the question:

If the point P is equidistant from M and N

then their distance between the points PM and PN must be equal.

Now, by using distance formula we will find distance between the two points.

Distance Between P and M

Distance formula

D=√(x₂- x₁)²+(y₂- y₁)²

M( -5,-2) and P(a,0)

x₁= -5 ;x₂ = a ; y₁= -2 & y₂ = 0

PM= √ (a+5)²+(0+2)²

identity :(a+b)²= a²+b²+2ab

PM= √ a²+5²+2(a)(5)+2²

PM= √a²+25+10a+4=√ a²+10a+29

Distance between P and N

P(a,0) N ( 3,2)

PN=√ (3-a)²+(2-0)²

PN= √ 9+a²-6a+4= √ a²-6a+13

Now , comparing both PM and PN

√ a²+10a+29=√a²-6a+13

Squaring both sides

PM²= PN²

(√a²+10a+29)²= (√a²-6a+13)²

a²+10a+29=a²-6a+13

a² gets cancelled as base same

=> 10a +29= -6a+13

=> 10a+6a=13-29

=> 16a = -16

=>a = -1

Thus,the point is (a,0) = (-1,0)

Therefore, (-1,0) is the point on the x-axis which is equidistant from the points M and N

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