Question

What is the perimeter of triangle ABC if the coordinates of the vertices are A(2,-9), B(2,21), and C(74,-9)?

Answer 1

See a solution process below:

Explanation:

To find the perimeter we need to find the length of each side using the formula for distance. The formula for calculating the distance between two points is:

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

Length of A-B:

`d_(A-B) = sqrt((color(red)(2) - color(blue)(2))^2 + (color(red)(21) - color(blue)(-9))^2)`

`d_(A-B) = sqrt((color(red)(2) - color(blue)(2))^2 + (color(red)(21) + color(blue)(9))^2)`

`d_(A-B) = sqrt((0)^2 + 30^2)`

`d_(A-B) = sqrt(0 + 30^2)`

`d_(A-B) = sqrt(30^2)`

`d_(A-B) = 30`

Length of A-C:

`d_(A-C) = sqrt((color(red)(74) - color(blue)(2))^2 + (color(red)(-9) - color(blue)(-9))^2)`

`d_(A-C) = sqrt((color(red)(74) - color(blue)(2))^2 + (color(red)(-9) + color(blue)(9))^2)`

`d_(A-C) = sqrt(72^2 + 0^2)`

`d_(A-C) = sqrt(72^2 + 0)`

`d_(A-C) = sqrt(72^2)`

`d_(A-C)= 72`

Length of B-C:

`d_(B-C) = sqrt((color(red)(74) - color(blue)(2))^2 + (color(red)(-9) - color(blue)(21))^2)`

`d_(B-C) = sqrt(72^2 + (-30)^2)`

`d_(B-C) = sqrt(5184 + 900)`

`d_(B-C) = sqrt(6084)`

`d_(B-C) = 78`

Perimeter of A-B-C:

`p_A-B-C = d_(A-B) + d_(A-C) + d_(B-C)`

`p_A-B-C = 30 + 72 + 78`

`p_A-B-C = 180`