Question

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

Answer 1

Step-by-step explanation:

let the point be (x,0)

(x+5)^2+4=(x-3)^2+4

x^2+10x+25=x^2-6x+9

10x+6x=9-25

16x=-16

x=-1

point is (-1,0)

Answer 2

Given:

• M (-5, -2)
• N (3, 2)

To find:

• Point which is equidistant from MN

Solution:

For points to be equidistant, the distance between the two points should be equal.

Let A(x, 0) be the point.

Now, we need to equate the distance between AM and AN.

Applying distance formula

AM = √[(x₂ - x₁)² + (y₂ - y₁)²]

⇒ AM = √(-5 - x)² + (-2 - 0)²

⇒ AM = √25 + x² + 10x + 4

⇒ AM = √x² + 10x + 29

AN = √[(x₂ - x₁)² + (y₂ - y₁)²]

➝ AN = √(3 - x)² + (2 - 0)²

➝ AN = √9 + x² - 6x + 4

➝ AN = √x² - 6x + 13

Equating AM and AN,

AM = AN

So, √(x² + 10x + 29) = √(x² - 6x + 13)

Squaring on both sides,

→ x² + 10x + 29 = x² - 6x + 13

Cancel x² on LHS and RHS

→ 16x = -16

→ x = -16 ÷ 16

→ x = -1

∴ The required point is (-1, 0)

Know more:

This question can also be solved by using midpoint formula that is

For x coordinate: (x₁ + x₂) ÷ 2 and

For y coordinate: (y₁ + y₂) ÷ 2