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?

3 Answers | Add Yours

electreto05's profile pic

electreto05 | College Teacher | (Level 1) Assistant Educator

Posted on

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.

ishpiro's profile pic

ishpiro | College Teacher | (Level 1) Educator

Posted on

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.

` `

` `

gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

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. 

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question