1)
The distance between points A and B is;
[tex]d= \sqrt{(x2-x1)^{2} + (y2-y1)^{2} } \\
d= \sqrt{(1-5)^{2} +(12+8)^{2}}\\
d= \sqrt{(-4)^{2} + (20)^{2}}\\
d=\sqrt{16+400}\\
d=\sqrt{416}\\
d=4\sqrt{26} [/tex],
and if you wanted to you could measure out the legs of that hypotenuse using the slope between A and B, but I'm going to show you the easier way;
Δx = 4
Δy = -20
So from the midpoint, add 4 to the x-value and subract 20 from the y-value.
{(5+4),(-8-20)}
(9,-28) is the endpoint.
2)
Do the same thing as I did above;
[tex]d= \sqrt{(x2-x1)^{2} + (y2-y1)^{2} } \\
d= \sqrt{(5-(-3))^{2} +(5-(-4))^{2}}\\
d=\sqrt{(5+8)^{2}+(5+4)^{2}}\\
d=\sqrt{(13)^{2}+(9)^{2}}\\
d=\sqrt{169+81}\\
d=\sqrt{250}\\
d=5\sqrt{10}[/tex]
So [tex]5\sqrt{10}[/tex] is the distance.
Sorry for the delay, life delayed me a ton ,:D
