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In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9 [#permalink]
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Here is a much simpler explanation and solution for this seemingly problem.
I prefer to use distance formula instead of height or triangle formulas. Simply, because it is easy to remember and apply too.

Just remember these 2 things:
1. All sides of Equilateral Triangle are equal
2. Distance formula: Under root[ (x-x1)^2 + (y-y1)^2 ]

Let's find the length of one side with distance formula:

(9)^2 + 0 = under root (81) = 9 {Side length}

Now here's something simple:
Apply distance formula to the coordinates of the point to be found (Keep coordinates: x,y)

Here is what you will get:
x^2 + y^2 = (9-x)^2 + y^2 {Distance formula from x,y to 0,0 and x,y to 9,0}
Solving this, cancel out y^2
Finally, you will get a value of x as 9/2 i.e. 4.5

This eliminates 2 options and now you can simply find y by substituting x as 9/2.

Apply this from x,y to 0,0 with distance as 9.
Solving it will come down to:
81/4 + y^2 = 81
Solve it y^2 = 243/4
Remove the square and y = 9.under root(3)/4
Hence the option (E)

Originally posted by KabxW on 19 Jul 2021, 07:58.
Last edited by KabxW on 22 Jul 2021, 02:11, edited 1 time in total.
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Re: In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9 [#permalink]
Hello from the GMAT Club BumpBot!

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Re: In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9 [#permalink]
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