Answer:
Step-by-step explanation:
Distance formula:
[tex]\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}[/tex]
Distance for AB
[tex]\sqrt{(4-2)^2+(1-(-5))^2}=\sqrt{4+36}=\sqrt{40}\\[/tex]
Distance for AC
[tex]\sqrt{(8-2)^2+(-3-(-5))^2}=\sqrt{36+4}=\sqrt{40}[/tex]
2.
same formula solve when positive
[tex]5=\sqrt{(6-3)^2+(y-2)^2}\\25=9+(y-2)^2\\16=(y-2)^2\\4=y-2\\6=y[/tex]
Now when it's negative
[tex]5=\sqrt{(6-3)^2+(y-2)^2}\\25=9+(y-2)^2\\16=(y-2)^2\\-4=y-2\\y=-2[/tex]
so y can equal 6 or -2
