Question

If the point A (x,2) is equidistant from the points B(8,-2) and C (2,-2), find the value of x .Also , find the length of AB.

Answer 1

Hello,

we have that:
AB=AC
so
AB²=AC²
then:
(x-8)²+(2+2)²=(x-2)²+(2+2)²;
x²-16x+64+16=x²-4x+4+16;
-16x+64=-4x+4;
-16x+4x=4-64;
-12x=-60;
12x=60;
x=60/12=5;
x=5

The value of x is 5

therefore, the point A(5,2) and B(8,-2);
then:
AB=√(5-8)²+(2+2)²=√(-3)²+(4)²=√9+16=√25= 5 

bye :-)

Answer 2

Answer:

The value of x is 5 and the length of AB is 5 units

Step-by-step explanation:

Given :The point A (x,2) is equidistant from the points B(8,-2) and C (2,-2)

To Find :  find the value of x

Since A is is equidistant from the points B  and C

SO, AB = AC

So, we will use distance formula : $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

So, $AB=\sqrt{(8-x)^2+(-2-2)^2$

$AC=\sqrt{(2-x)^2+(-2-2)^2$

So , $\sqrt{(2-x)^2+(-2-2)^2}=\sqrt{(8-x)^2+(-2-2)^2}$

$(2-x)^2=(8-x)^2$

$4+x^2-4x=64+x^2-16x$

$4-4x=64-16x$

$12x=60$

$x=5$

So, $AB=\sqrt{(8-5)^2+(-2-2)^2}=5$

Hence the value of x is 5 and the length of AB is 5 units