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What is the perimeter of AABC with vertices A(-2,-3), B(6,-3), and C(-2,3)?

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Notice that we are given the three vertices of a triangle:

A = (-2, -3)

B = (6, -3)

C = (-2, 3)

The perimeter is simply the addition of the length of each side.

So let's start with the "easy" sides which are those whose vertices share a same coordinate, since it is easier to find the distance of two points that share the same line in the coordinate system.

For example: Points A and B share the y-oordinate "-3" so they are located at the same level in the y-axis. Therefore the distance between these two is simply the distance between 6 and -2 in the x axis = 6 - (-2) = 8

The other two vertices that share the same coordinate are: A and C (they share the x-coordinate "-2" Therefore they are located at the same x value and the distance between them is simply the distance between the y-values: 3 - (-3) = 6

now, we have to find the distance between the two vertices that don't share any coordinate. B and C.

That distance is:


\sqrt[]{(-3-3)^2+(6--2)^2}=\sqrt[]{6^2+8^2}=\sqrt[]{36+64}=\sqrt[]{100}=10

Then the third side has a length of 10 units.

Therefore, the perimeter of the triangle is: 8 + 6 + 10 = 24 units

User Hj David Kearns
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