Question

Show that the point (1,2) is equidistant from the points (-2,-2)(4,6) and (5,5)

Answer 1

Given:

Four points

Let A(1, 2)

B(-2,-2)

C(4,6) and D(5,5)

To prove:

A is equidistant from B, C and D.

Solution:

Here, we need to find the distances AB, AC and AD and then compare their values.

First of all, let us learn about the distance formula:

Distance formula :

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where $(x_1, y_1), (x_2, y_2)$ are the two points for which the distance is to be calculated.

First, let us consider A and B:

$x_2 = -2\\x_1 = 1\\y_2 = -2\\y_1 = 2$

So, the distance AB:

$AB = \sqrt{(-2-1)^2+(-2-2)^2}\\\Rightarrow AB = \sqrt{(-3)^2+(-4)^2}\\\Rightarrow AB = \sqrt{9+16} = \sqrt{25}\\\Rightarrow \bold{AB=5 \ units}$

Now, let us consider A and C:

$x_2 = 4\\x_1 = 1\\y_2 = 6\\y_1 = 2$

So, the distance AC:

$AC = \sqrt{(4-1)^2+(6-2)^2}\\\Rightarrow AC = \sqrt{(3)^2+(4)^2}\\\Rightarrow AC = \sqrt{9+16} = \sqrt{25}\\\Rightarrow \bold{AC=5 \ units}$

Now, let us consider A and C:

$x_2 = 5\\x_1 = 1\\y_2 = 5\\y_1 = 2$

So, the distance AD:

$AD = \sqrt{(5-1)^2+(5-2)^2}\\\Rightarrow AD = \sqrt{(4)^2+(3)^2}\\\Rightarrow AD = \sqrt{16+9} = \sqrt{25}\\\Rightarrow \bold{AD=5 \ units}$

Now, comparing the values of AB, AC and AD:

We can see that AB = AC = AD

Hence proved that:

the point (1,2) is equidistant from the points (-2,-2)(4,6) and (5,5)