Respuesta :
the perimeter of any figure, is all its sides summed up
thus
[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) P&({{ -2}}\quad ,&{{9}})\quad % (c,d) Q&({{ 7}}\quad ,&{{ -3}})\\ Q&({{ 7}}\quad ,&{{ -3}})\quad % (c,d) R&({{ -2}}\quad ,&{{ -3}})\\ R&({{ -2}}\quad ,&{{ -3}})\quad % (c,d) P&({{ -2}}\quad ,&{{ 9}})\\ \end{array}\ \ % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}[/tex]
[tex]\bf PQ=\sqrt{[7-(-2)]^2+[-3-9]^2}\implies PQ=\sqrt{(7+2)^2+(-3-9)^2} \\\\\\ PQ=\sqrt{81+144}\\\\ -------------------------------\\\\ QR=\sqrt{[-2-7]^2+[-3-(-3)]^2}\implies QR=\sqrt{(-9)^2+(-3+3)^2} \\\\\\ QR=\sqrt{81}\\\\[/tex]
[tex]\bf -------------------------------\\\\ RP=\sqrt{[-2-(-2)]^2+[9-(-3)]^2} \\\\\\ RP=\sqrt{(-2+2)^2+(9+3)^2}\implies RP=\sqrt{144}[/tex]
so, the perimeter is PQ + QR + RP
simplify them, sum them up, and that's the perimeter of the triangle.
thus
[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) P&({{ -2}}\quad ,&{{9}})\quad % (c,d) Q&({{ 7}}\quad ,&{{ -3}})\\ Q&({{ 7}}\quad ,&{{ -3}})\quad % (c,d) R&({{ -2}}\quad ,&{{ -3}})\\ R&({{ -2}}\quad ,&{{ -3}})\quad % (c,d) P&({{ -2}}\quad ,&{{ 9}})\\ \end{array}\ \ % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}[/tex]
[tex]\bf PQ=\sqrt{[7-(-2)]^2+[-3-9]^2}\implies PQ=\sqrt{(7+2)^2+(-3-9)^2} \\\\\\ PQ=\sqrt{81+144}\\\\ -------------------------------\\\\ QR=\sqrt{[-2-7]^2+[-3-(-3)]^2}\implies QR=\sqrt{(-9)^2+(-3+3)^2} \\\\\\ QR=\sqrt{81}\\\\[/tex]
[tex]\bf -------------------------------\\\\ RP=\sqrt{[-2-(-2)]^2+[9-(-3)]^2} \\\\\\ RP=\sqrt{(-2+2)^2+(9+3)^2}\implies RP=\sqrt{144}[/tex]
so, the perimeter is PQ + QR + RP
simplify them, sum them up, and that's the perimeter of the triangle.
