# How do you find the perimeter of a triangle on a coordinate plane whose points are A (-3, 6), B (-3, 2), C (3, 2)?

Jun 7, 2015

Be a triangle formed with the points A[-3;6], B[-3;2] and C[3;2].

The perimeter of this triangle is

$\Pi = A B + B C + A C$

In a plane, the distance between two points M and N is given by

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

Therefore

$A B = \sqrt{{\left(- 3 - \left(- 3\right)\right)}^{2} + {\left(6 - 2\right)}^{2}} = \sqrt{0 + {4}^{2}} = 4$

$B C = \sqrt{{\left(- 3 - 3\right)}^{2} + {\left(2 - 2\right)}^{2}} = \sqrt{{\left(- 6\right)}^{2} + 0} = 6$

$A C = \sqrt{{\left(- 3 - 3\right)}^{2} + {\left(6 - 2\right)}^{2}} = \sqrt{{\left(- 6\right)}^{2} + {4}^{2}} = \sqrt{52}$

$\rightarrow \Pi = 4 + 6 + \sqrt{52} = 10 + \sqrt{52} = 10 + 2 \sqrt{13}$