What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?14.6 units15.5 units21.0 units21.6 units

see the attached figure to better understand the problemwe know that the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]Let [tex]A(-3,3)\ B(3,4))\ C(3,-3)[/tex]  Step 1Find the distance AB[tex]A(-3,3)\ B(3,4)[/tex]  substitute in the formula[tex]d=\sqrt{(4-3)^{2}+(3+3)^{2}}[/tex][tex]d=\sqrt{(1)^{2}+(6)^{2}}[/tex][tex]dAB=\sqrt{37}\ units[/tex]Step 2Find the distance BC[tex]B(3,4))\ C(3,-3)[/tex]  substitute in the formula[tex]d=\sqrt{(-3-4)^{2}+(3-3)^{2}}[/tex][tex]d=\sqrt{(-7)^{2}+(0)^{2}}[/tex][tex]dBC=7\ units[/tex]Step 3Find the distance AC[tex]A(-3,3)\ C(3,-3)[/tex]  substitute in the formula[tex]d=\sqrt{(-3-3)^{2}+(3+3)^{2}}[/tex][tex]d=\sqrt{(-6)^{2}+(6)^{2}}[/tex][tex]dAC=\sqrt{72}\ units[/tex]Step 4Find the perimeter of the trianglewe know thatthe perimeter of the triangle is the sum of the length sides of the triangle[tex]P=AB+BC+AC[/tex]substitute the values[tex]P=\sqrt{37}\ units+7\ units+\sqrt{72}\ units=21.6\ units[/tex]thereforethe answer isthe perimeter of the triangle is [tex]21.6\ units[/tex]