# How do you find the perimeter of the triangle whose vertices are these specific points on a plane (-1,6) (4,-5) (-2, -4)?

Oct 20, 2015

A little over $28$

#### Explanation:

The perimeter is going to be equal to the sum of the lengths of the sides, which can be found by using the distance formula;

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Plugging in the points for the first two vertices, we get;

${d}_{1} = \sqrt{{\left(4 + 1\right)}^{2} + {\left(- 5 - 6\right)}^{2}} = \sqrt{146} \cong 12$

For the second two;

${d}_{2} = \sqrt{{\left(- 2 - 4\right)}^{2} + {\left(- 4 + 5\right)}^{2}} = \sqrt{37} \cong 6$

And for the last pair;

${d}_{3} = \sqrt{{\left(- 2 + 1\right)}^{2} + {\left(- 4 - 6\right)}^{2}} = \sqrt{109} \cong 10$

So the sum of the three sides is roughly;

$12 + 6 + 10 = 28$

You can get a more accurate decimal approximation by using a calculator to evaluate the square root values above instead of approximating them.