Question

Please click the photo and there are two questions to solve .Please help me guys

Answer 1

Answer:

Step-by-step explanation:

Distance formula:

$\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}$

Distance for AB

$\sqrt{(4-2)^2+(1-(-5))^2}=\sqrt{4+36}=\sqrt{40}$

Distance for AC

$\sqrt{(8-2)^2+(-3-(-5))^2}=\sqrt{36+4}=\sqrt{40}$

2.

same formula solve when positive

$5=\sqrt{(6-3)^2+(y-2)^2}\\25=9+(y-2)^2\\16=(y-2)^2\\4=y-2\\6=y$

Now when it's negative

$5=\sqrt{(6-3)^2+(y-2)^2}\\25=9+(y-2)^2\\16=(y-2)^2\\-4=y-2\\y=-2$

so y can equal 6 or -2