Question

Determine whether the three points are the vertices of a right triangle. (6, -6), (9, -6), (9, 1)

Answer 1

Answer:

The three given points are the vertices of a right triangle.

Step-by-step explanation:

To determine that the three points are the vertices of a right triangle let us find the distance between each two points

The formula of the distance is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

∵ $x_{1}$ = 6 and $y_{1}$ = -6

∵ $x_{2}$ = 9 and $y_{2}$ = -6

∴ $d_{1}$ = $\sqrt{(9-6)^{2}+(-6--6)^{2}}=\sqrt{9+0}$

∴ $d_{1}$ = 3

∵ $x_{1}$ = 6 and $y_{1}$ = -6

∵ $x_{2}$ = 9 and $y_{2}$ = 1

∴ $d_{2}$ = $\sqrt{(9-6)^{2}+(1--6)^{2}}=\sqrt{9+49}$

∴ $d_{2}$ = $\sqrt{58}$

∵ $x_{1}$ = 9 and $y_{1}$ = 1

∵ $x_{2}$ = 9 and $y_{2}$ = -6

∴ $d_{3}$ = $\sqrt{(9-9)^{2}+(-6-1)^{2}}=\sqrt{0+49}$

∴ $d_{3}$ = 7

Let us use the fact that if the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle at the vertex which opposite to the longest side

∵ The longest side is $\sqrt{58}$

∵ The other two sides are 3 and 7

∵ ( $\sqrt{58}$ )² = 58

∵ (3)² + (7)² = 9 + 49 = 58

∴  ( $\sqrt{58}$ )² = (3)² + (7)²

- By using the fact above

∴ The triangle is a right triangle

The three given points are the vertices of a right triangle.