Vertices of a triangle?

[wrongnmbr]

Can't quite figure this one out. I'm given the points (-1,0) (-13,-12). Now I'm supposed to find a third point so that the three points form the vertices of an isosceles triangle.

Which of these points is accurate:
(3, -17)
(-16, -15)
(2, 3)
(-7, -6)
(2, -15)

I know I'm supposed to use the distance formula d = ?(-1 + 13)^2 + (0 + 12)^2 and I come up with ?288, but I'm lost after this point. Where do I go from here? Any help would be appreciated!

 

[mmm4444bot]

Have you learned that two of the three triangle sides must be equal ? (That's the definition of an isosceles triangle.)

Your calculation for the given side is correct.

Next, beginning with the first multiple choice (3, -17), calculate the lengths of the remaining two sides.

If two out of the three sides are equal, you're done. OOPS. I misspoke.

You're not done until you've checked each of the multiple choices because more than one of the given choices works. I wrongly assumed that only one of the multiple-choice answers is possible. I should have checked them all the first time because it's so quick and easy.

 

[galactus]

This one is not among your choices, but it works.

The point (-1,0) is the top of the triangle. The point (-13,-12) is the left corner.

We then know that the right side of the triangle has to have the same length as the left side, which is $\displaystyle 12\sqrt{2}$

The lower right corner has cooridnates (x,-12)

So, by the distance formula, the distance between the peak at (-1,0) and (x,-12) is the same as the other side, $\displaystyle 12\sqrt{2}$.

Therefore:

$\displaystyle \sqrt{(-1-x)^{2}+(0-(-12))^{2}}=12\sqrt{2}$

$\displaystyle (-1-x)^{2}+144=288$

$\displaystyle x^{2}+2x-143=0$

Two solutions. One is (-13,-12) and the other is the x in (x,-12) which forms the right lower vertex needed to form a isosceles

 

[BigGlenntheHeavy]

\(\displaystyle Question: If the distance from \ (-1,0) \ to \ (-13,-12), \ = \ 12\sqrt2 \ is \ the \ base \ of \ the \ isosceles \ triangle, \\)

$\displaystyle how \ many \ isosceles \ triangles \ could \ I \ have? \ and \ if \ 12\sqrt2 \ is \ the \ length \ of \ one \ of \ the \ sides$

$\displaystyle of \ the \ isosceles \ triangles, \ then \ how \ many \ of \ them \ could \ I \ have?$

$\displaystyle Note: If 12\sqrt2 \ is \ the \ base, \ then \ its \ perpendicular \ bisector \ divides \ the \ base \ into \ an \ infinite$

$\displaystyle number \ of \ isosceles \ triangles.$

$\displaystyle And \ if \ 12\sqrt2 \ is \ one \ of \ the \ sides, \ then \ let \ 12\sqrt2 \ be \ the \ radius \ of \ a \ circle \ and \ again$

$\displaystyle we \ have \ an \ infinite \ number \ of \ isosceles \ triangles.$

$\displaystyle Note: \ One \ infinity \ is \ open \ and \ one \ infinity \ is \ closed. \ Now, \ according \ to \ Cantor, \ are \ these \ two$

$\displaystyle infinities \ equal?$