# Thread: confused with these trigono....

1. ## confused with these trigono....

Show that the given points are the vertices of an isosceles triangle. Use the

distance formula. (Show that two sides of the triangle are equal in length).

1.) A (-1, -6), B (2, -7), C (-2, -3)

2.) A (-9, 3), B (-4, 8), C (0, -1)

3.) A (2, -3), B (3, 4), C (-5, -2)

4.) A (-3, 0), B (-6, 7), C (1, 4)

2. Originally Posted by orochimaru700
Show that the given points are the vertices of an isosceles triangle. Use the

distance formula. (Show that two sides of the triangle are equal in length).

1.) A (-1, -6), B (2, -7), C (-2, -3)

2.) A (-9, 3), B (-4, 8), C (0, -1)

3.) A (2, -3), B (3, 4), C (-5, -2)

4.) A (-3, 0), B (-6, 7), C (1, 4)
This seems pretty straightforward. What part don't you understand? Do you know the distance formula?

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

3. don't know how to solve for the following, our instructor said, that we plot the following and solve for each problem... don't know what to do

4. Originally Posted by orochimaru700
don't know how to solve for the following, our instructor said, that we plot the following and solve for each problem... don't know what to do
The question tells you exactly what to do.

You work out the distance between the vertices using the distance formula.

Once you have all three distances of the triangles, you should be able to see that in each triangle, two distances are equal. So the triangles are isosceles.

E.g. for the first one, you have

$A = (-1, -6), B = (2, -7), C= (-2, -3)$.

$AC = \sqrt{[-2 - (-1)]^2 + [-3 - (-6)]^2}$

$= \sqrt{(-2 + 1)^2 + (-3 + 6)^2}$

$= \sqrt{(-1)^2 + 3^2}$

$= \sqrt{1 + 9}$

$= \sqrt{10}$.

Now work out the distance of $AB$ and $BC$.

5. ok thanks for the first one, i'll try to solve the other problem...