MATHSS!!! Triange ABC....?

Question

Triangle ABC has a right angle at B. The vertices are A(-2,9) B(2,8) and C(1,z) What is the value of z.

Answer 1

Phythagoren theorem:

If a triangle is a right angle triangle, then (Hypotenuse)^2 = (Side)^2 + (Side)^2.

Here, tringle ABC has a right angle at B.

So, AC^2 = AB^2 + BC^2.

The points are A(- 2, 9), B(2, 8) , and C(1, z).

AC = distance between the points A(- 2, 9) and C(1, z) = √[ (1 + 2)^2 + (z - 9)^2 ] = √[3^2 + (z - 9)^2 ] = √(9 + (z - 9)^2).

BC = distance between the points B(2, 8) and C(1, z) = √[ (1 - 2)^2 + (z - 8)^2 ] = √[(- 1)^2 + (z - 8)^2 ] = √(1 + (z - 8)^2).

AB = distance between the points A(- 2, 9) and B(2, 8) = √[ (2 + 2)^2 + (8 - 9)^2 ] = √[(4)^2 + (- 1)^2 ] = √16 + 1 = √17.

AC^2 = AB^2 + BC^2.

[ √(9 + (z - 9)^2) ]^2 = [√17]^2 + [ √(1 + (z - 8)^2) ]^2

9 + (z - 9)^2 = 17 + 1 + (z - 8)^2

9 + z^2 - 18z + 81 = 18 + z^2 - 16z + 64

90 - 18z = 82 - 16z

2z = 90 - 82 = 8

⇒ z = 8/2 = 4.

Therefore, the value of z = 4.