# For the given verticies, determine whether or not ABC is a right triangle A(13,3), B (3,4) and C(-2,-20) I am not sure how to approach this... any ideas?

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One way to do this is by checking if the lengths of the sides of this triangle satisfy the Pythagorean Theorem: `a^2 + b^2 = c^2` .

To find the lengths of the sides, use the distance formula:

`d = sqrt((y_2-y_1)^2 + (x_2- x_1)^2)`

So, the length of the side AB will be

`AB = sqrt((4 - 3)^2 + (3 - 13)^2) = sqrt(1 + 100) = sqrt(101)`

The length of the side BC will be

`BC = sqrt((-20 - 4)^2 + (-2-3)^2) = sqrt(576 + 25) = sqrt(601)`

The length of the side AC will be

`AC = sqrt(-20-3)^2 + (-2 - 13)^2) = sqrt(529 + 225)=sqrt(754) `

It can be seen that the side AC is the longest, so it this was a right triangle, it would be the hypotenuse and the relationship

`AB^2 + BC^2 = AC^2` would be satisfied. However, the squares of the lengths of the sides AB and BC, 101 +601 = 702 is not equal to the square of AC, 754, so

**ABC is NOT a right triangle.**

Another way to approach this problem is to calculate the slopes of the lines AB, BC and AC and check if there is a pair of two slopes that are negative reciprocals of each other. If there is, then ABC is a right triangle.

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We know that two lines are perpendicular if their slopes are mutually reciprocal. Two lines with slopes m1 and m2 are perpendicular if they meet the following:

m1 = -1/m2

With this in mind and, knowing the coordinates of the vertices; we just have to find the slopes of the straight lines connecting the vertices and see if they meet the above relationship.

The slope of a line is:

m = (y2 - y1)/(x2 - x1)

The coordinates of the points:

A(13; 3), B (3; 4) and C(-2; -20)

For line through AB, we have:

mAB = (4 – 3)/(3 – 13) = -1/10

For line through BC:

mBC = (-20 – 4)/(-2 – 3) = 24/5

For line through CA:

mCA = (-20 – 3)/(-2 – 13) = 23/15

Comparing, the obtained values of m, we see that do not meet the condition of perpendicularity, expressed at the beginning; **so that the triangle does not have a right angle.**

For the triangle ABC, lets first calculate the length of sides:

AB = sqrt[(13-3)^2 + (3-4)^2)] = sqrt[10^2 + 1^2] = sqrt(101) = 10.05

BC= sqrt[(3--2)^2+(4--20)^2] = sqrt[5^2 + 24^2] = sqrt(601) = 24.52

CA = sqrt[(-2-13)^2+(-20-3)^2] = sqrt[15^2 + 23^2] = sqrt(754) = 27.46

Now for a right angled triangle, square of hypotenuse is equal to sum of square of the other sides.

so, for ABC to be a right triangle, CA^2 = AB^2 + BC^2

CA^2 = 754

AB^2+BC^2 = 101+601 = 702

Since LHS is not equal to RHS, ABC is not a right angled triangle.