Is the triangle whose vertices are the points A(0 , -1) , B(2 , 1) and C(-4 , 3) a right triangle?

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The triangle ABC has vertices at A(0, -1) , B(2 , 1) and C(-4, 3).

The length of the sides is

AB = sqrt [(2 - 0)^2 + (1 + 1)^]

=> sqrt (4 + 4) = sqrt 8

BC = sqrt[(2 + 4)^2 + (3 -1)^2]

=> sqrt( 36 + 4) = sqrt 40

CA = sqrt [( 4 - 0)^2 + ( 3 + 1)^2]

=> sqrt (16 + 16) = sqrt 32

As CA^2 + AB^2 = BC^2, by the Pythagorean theorem we know that the triangle is a right triangle with hypotenuse BC.

**The triangle ABC is a right triangle.**

To determine is the the triangle is a right angle, we will need to prove that two sides are perpendicular.

Let us calculate the slopes of each of the sides.

The slope for AB = ( 1+1)/2-0) = 2/2 = 1

The slope for AC = ( 3+1)/(-4-0) = 4/-4 = -1

The slope for BC = ( 3-1)/(-4-2) = 2/-6 = -1/3

If two line are perpendicular, then we know that the product of the slopes is -1.

We notice that the slope for AB * the slope for AC = -1

Then the lines AB and AC are perpendicular.

**Then we conclude that the vertices's form a right angle triangle at A.**

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