Answer: P=7+3√5+2√13 units S=21 quadratic units
Step-by-step explanation:
[tex]A(1,-2)\ \ \ \ B(1.5)\ \ \ \ C(7,2)\\[/tex]
The perimeter of a triangle
[tex]|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2} \\|AB|=\sqrt{(1-1)^2+(5-(-2))^2} \\|AB|=\sqrt{0^2+(5+2)^2}\\|AB|=\sqrt{0+7^2}\\ |AB|= \sqrt{7^2} \\|AB|=7[/tex]
[tex]|BC|=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2} \\|BC|=\sqrt{(7-1)^2+(2-5)^2}\\ |BC|=\sqrt{6^2+(-3)^2}\\ |BC|=\sqrt{36+9}\\ |BC|= \sqrt{45} \\|BC|=\sqrt{9*5}\\ |BC|=3\sqrt{5}[/tex]
[tex]|AC|=\sqrt{(x_C-x_A)^2+(y_C-x_A)^2}\\ |AC|=\sqrt{(7-1)^2+(2-(-2)^2} \\|AC|=\sqrt{6^2+(2+2)^2}\\ |AC|=\sqrt{36+4^2}\\ |AC|=\sqrt{36+16} \\|AC|=\sqrt{52}\\ |AC|=\sqrt{4*13} \\|AC|=2\sqrt{13} \\[/tex]
[tex]Thus,\\\\P_{ABC}=|AB|+|BC|+|AC|\\\\P_{ABC}=7+3\sqrt{5}+2\sqrt{13}\ units[/tex]
The area of a triangle
Taking A as the first vertex, we find:
[tex]\displaystyle\\Find\ |\overline{AB}*\overline{AC}|\\\left|\begin{array}{cc}x_A-x_C&y_A-y_C\\x_B-x_C&y_B-y_C\\\end{array}\right|=\\\\\left|\begin{array}{ccc}1-7&-2-2\\1-7&5-2\\\end{array}\right| =\\\\\left|\begin{array}{ccc}-6&-4\\-6&3\\\end{array}\right| =\\\\|(-6)*3-(-6)*(-4)|=\\\\|-18-24|=\\\\|-42|=\\\\42\\\\S_{ABC}=\frac{1}{2} |\overline{AB}*\overline{AC}|\\\\S_{ABC}=\frac{1}{2}(42)\\\\S_{ABC}=21 \ quadratic\ units[/tex]
