Find all possible values for y

Question

so that (5,8),(-4,11) and (2,y) are the vertices of a right triangle.?

Answer 1

Let the vertices be A = (5, 8), B (-4, 11) and C (2, y).

In right angle triangle AC² = AB² + BC² by pythagorus theorem.

Here AC, AB, BC are the distances between two vertices.

Therefore AC²  = [√ [(5-2)² + (8-y)²]]²  = (3)² + (64 -16y + y²) = 9 + 64 -16y + y² = y² - 16y + 73.

AB²  = [√ [(5+4)² + (8-11)²]]² = (9)² + (3)² = 81 + 9 = 90.

BC²  = [√ [(-4-2)² + (11-y)²]]²  = (-6)² + (121 - 22y + y²) = 36 + 121 - 22y + y² = y² - 22y + 157.

AC² = AB² + BC²

y² - 16y + 73 = 90 + y² - 22y + 157

y² - 16y + 73 - 90 - y²  + 22y - 157 = 0

 - 16y + 73 - 90 + 22y - 157 = 0

6y + 73 - 247 = 0

6y - 174 = 0

6y = 174

y = 174/6

y = 29

Therefore y = 29.