# Math Help - the distance and midpoint formulas

1. ## the distance and midpoint formulas

Sketch a diagram for each problem on the grid provided, then solve the problem.

12. Show that the triangle with vertices (0,0), (6,0) and (3,3) is an isosceles right triangle-that is, a right triangle with two sides of the same length.

how do i solve this?

2. Originally Posted by comet2000
Sketch a diagram for each problem on the grid provided, then solve the problem.

12. Show that the triangle with vertices (0,0), (6,0) and (3,3) is an isosceles right triangle-that is, a right triangle with two sides of the same length.

how do i solve this?
Everyone wave to our good friend, the distance formula

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

You can use that formula to find out the length of the sides, if two sides are the same, then the triangle is isosceles, if you need more help understanding finding the lengths then you can just ask.

Now how do you find out if it's a right triangle? To do that, we can use an even better friend of mathematics, the pythagorean theorum.

$a^2 + b^2 = c^2$

You have to find the lengths of the sides, and then substitute them in, if the equation is correct, then you have a right triangle.

3. here the given points are A(0,0) B(3,3) C(6,0)
Distance between AB=sqrt(9+9)=sqrt(18)
distance between Ac=sqrt(36)=6
Distance between BC=sqrt(9+9)=sqrt(18)
clearly here any two sides are equal which is isoscles of a traingle

4. Originally Posted by tutor
here the given points are A(0,0) B(3,3) C(6,0)
Distance between AB=sqrt(9+9)=sqrt(18)
distance between Ac=sqrt(36)=6
Distance between BC=sqrt(9+9)=sqrt(18)
clearly here any two sides are equal which is isoscles of a traingle
Moreover, $\left( {\sqrt {18} } \right)^2 + \left( {\sqrt {18} } \right)^2 = \left( 6 \right)^2$

5. huh? i don't understand what you mean.

use the pythagorean theorum and distance formula to solve the problem?