Question

find the perimeter of a triangle with vertices (2,4),(1,3),and (2,-3)

Answer 1

A(2,4) and B(1,3)
x2=1 x1=2 y2=3 y1=4
So,
(1-2)^2 + (3-4)^2 = 1+1=2 square root
B(1,3) and C(2,-3)
so,
(2-1)^2 + (-3-3)^2 = 1+36=37 square root.
A(2,4) and C(2,-3)
(2-2)^2 + (-3-4)^2 = 0 + 49 = square root of 49 is 7 so
Sides are
\[7, \sqrt{37}, \sqrt{2}\]
For the perimeter, just add them:
\[\sqrt{37}+\sqrt{2}+7\] units.

Answer 2

why do you do the square roots?

Answer 3

\[\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}}\]
This is the formula for finding the length of a side, so:
\[\sqrt{(1-2)^{2}+(3-4)^{2}}\]
\[\sqrt{1+1}=\sqrt{2}\]
That's how we get the square roots :)

Answer 4

ohh okay thank you!

Answer 5

My pleasure! :)