Answer: AB≈10,8, C(-1;0,5).
Step-by-step explanation:
4.
[tex](-2,-2)\ \ \ \ (8,-6)\ \ \ \ \overline {AB}=?\\[/tex]
Equation of a straight line
[tex]\boxed {\overline {AB}=\sqrt{(x_B+x_A)^2+(y_B-y_A)^2} }[/tex]
[tex]x_A=-2\ \ \ \ x_B=8\ \ \ \ y_A=-2\ \ \ \ \ y_B=-6\\\\\overline{AB}=\sqrt{(8-(-2))^2+(-6-(-2))^2} \\\\\overline{AB}=\sqrt{(8+2)^2+(-6+2)^2} \\\\\overline{AB}=\sqrt{10^2+(-4)^2} \\\\\overline{AB}=\sqrt{100+16} \\\\\overline{AB}=\sqrt{116} \\\\\overline{AB}\approx10,8.[/tex]
5.
[tex](-2,-2)\ \ \ \ (0,3)\ \ \ \ the\ midpoint\ C\ of\ AB=?\\[/tex]
Line Midpoint Equation
[tex]\boxed {C_x=\frac{X_A+X_B}{2} }}\ \ \ \boxed {C_y=\frac{Y_A+Y_B}{2} }[/tex]
[tex]X_A=-2\ \ \ \ \ X_B=0\ \ \ \ \ Y_A==-2\ \ \ \ Y_B=3\\\\C_x=\frac{-2+0}{2} \\\\C_x=\frac{-2}{2} \\\\C_x=-1.\\\\C_y=\frac{-2+3}{2}\\\\C_y=\frac{1}{2}\\\\ C_y=0,5.\\\\Hence,\ \ \ \ C_{AB}(-1;0,5).[/tex]
