A triangle that has two sides of equal length is called isosceles. Make up an example of an isosceles triangle, one of whose vertices is (3,5). If you can, find a triangle that does not have any horizontal or vertical sides.
@agent0smith
I mainly need help with finding a triangle that does not have horizontal or vertical sides. What does that mean?
that seems logical
|dw:1473617311980:dw| Distance between (3, 5) and (a, 0) must equal distance between (3, 5) and (c, d) for it to be isosceles. Pick a value of \(a\), then you can work on finding \(c\) and \(d\) using distance formula. You'll be able to also pick a reasonable value of \(c\), once you've found the (3, 5) and (a, 0) distance.
Could I do something with the pythagorean theorem?
Pick a value of a, then find the distance between (3, 5) and (a, 0).
ok
i chose that a =1 and i had the distance of \(\sqrt{29}\)
Now pick a reasonable value of c (it must be somewhat close to x=3, within sqrt29 units, or else the triangle won't be isosceles) Distance between (3, 5) and (c, d) will equal to sqrt 29, solve for d.
how did you figure out that the value of c needs to close to x = 3?
Draw it and find out. If you make it too far away, will it be isosceles?
I am confused. If x = 3, it would be right underneath the point (3,5).
I chose x = 5
You said you didn't want horizontal or vertical lines, so why would you pick x=3?
Distance between (3, 5) and (c, d) will equal to sqrt 29, solve for d.
\(\sqrt{29} = \sqrt{(3-5)^2 + (5-d)^2}\)
\[(\sqrt{29})^2 = (\sqrt{3-5)^2 + (5-d)^2})^2 \] \(29 = 4 + (5-d)^2\)
Then I get that d = 0, 10
Yep.
But how do I know which one to choose from?
Well, they both work to make a triangle...but you didn't want vert/horiz lines... so you should be able to choose based on that.
If i have y = 0 then i don't think that it would work because it would have horizontal and vertical sides. I guess 10?
would that be correct @agent0smith ?
Sure. Draw them and it should look isosceles.
Thank you! Can you help me with another question please?
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