Question

Find the point on x- axis which
equidistance form (2,-2) & (-4,2).​

Answer 1

Answer:

-3

Step-by-step explanation:

as the point on x axis are equidistant from (2,-2) and (-4,2), so first I am going to find the distance of the two points (x,y) and(2,-2).

(x-2)^2+(y+2)^2

=x^2-4x+4+y^2+4x+4

=x^2+y^2+8

so, the distance between the two points is

$\sqrt{(x { }^{2} + y {}^{2} + 8) } > > > 1$

similarly the distance between another two points (x,y) and (-4,2) is

$\sqrt{(x ^{2} + y ^{2} + 4x + 20) } > > > 2$

as they are equidistant from(2,-2) and (4,-2) so,

1no distance=2 no distance,

after doing this equation I found x=-3

if this solve is wrong then plz inform me

Answer 2

Answer:

• Midpoint = (-1, 0)

Step-by-step explanation:

Let the points be:

• A (2, -2)
• B (-4, 2)

Given:

2 coordinates A and B are given.

• x₁ = 2
• x₂ = -4
• y₁ = -2
• y₂ = 2

To find:

We are supposed to find the midpoint A and B.

Solution:

Applying midpoint formula,

For X coordinate,

AB = (x₁ + x₂) ÷ 2

⇒ AB = (2 - 4) ÷ 2

⇒ AB = -1

For Y coordinate,

AB = (y₁ + y₂) ÷ 2

⇒ AB = (-2 + 2) ÷ 2

⇒ AB = 0 ÷ 2

⇒ AB = 0

∴ Midpoint = (-1, 0)

KNOW MORE:

This question can also be solved by applying distance formula i.e

AB = √[(x₂ - x₁)² + (y₂ - y₁)²]