Question

Буду очень благодарна!

Answer 1

$A(1;2) \; ,\; B(-1;2)\; ,\; C(3;0)\\\\1)\quad AB=\sqrt{(-1-1)^2+(2-2)^2}=2$

$AC=\sqrt{(3-1)^2+(0-2)^2}=\sqrt{4+4}=2\sqrt2\\\\BC=\sqrt{(3+1)^2+(0-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt5\\\\P=AB+AC+BC=2+2\sqrt2+2\sqrt5=2(1+\sqrt2+\sqrt5)$

$2)\quad AB:\; \; y_{A}=y_{B}\; \; \; \to \; \; \underline {y=2}\\\\BC:\; \; \frac{x-x_{B}}{x_{C}-x_{B}} = \frac{y-y_{B}}{y_{C}-y_{B}} \\\\ \frac{x+1}{3+1} = \frac{y-2}{0-2} \; ,\; \; \frac{x+1}{4} = \frac{y-2}{-2} \; \; \to \; \; -2(x+1)=4(y-2)\; |:(-2)\\\\x+1=-2(y-2)\\\\x+1=-2y+4\\\\x+2y-3=0\\\\2y=-x+3\\\\\underline {y=-0,5x+1,5}$