In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9

Question

In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9, 0). What could be the coordinates of the third vertex?

(A) (0, 4.5)

(B) (4.5, 4.5)

(C)  (\frac{9\sqrt{3}}{2}, \ \frac{9\sqrt{3}}{2})

(D)  (4.5, \ 9\sqrt{3})

(E)  (4.5, \ \frac{9\sqrt{3}}{2})

Kinshook · Answer

Given: In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9, 0). 
Asked: What could be the coordinates of the third vertex?

Side of the equilateral triangle = 9
Height of the equilateral triangle =  \sqrt{81 - 81/4} = 9\sqrt{3}/2

Coordinates of the third vertex =  (4.5, 9\sqrt{3}/2)

IMO E

KabxW · Answer

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

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)

Fdambro294 · Answer

If you look at all the answers choices, they all appear in quadrant 1.

Since the one side lies down along the X Axis and is 9 units (from (0 , 0) to (9 , 0))

the Height of the equilateral triangle will extend vertically from the X Axis into Quadrant 1

Rule: from any vertex, the Height = perpendicular Bisector = median = angle Bisector

Rule: the Height for a particular given equilateral triangle is given by——->

Height = (side) * sqrt(3) * (1/2)

Since the Height will also be the Median and perpendicular Bisector of the 9 unit side along the X Axis, the X coordinate of the 3rd unknown Vertex should be at the midpoint of the 2 Given vertices’ X Coordinates:

(9 + 0) / 2 = 4.5 = X Coordinate

Furthermore, from this X coordinate, we can find vertex’s accompanying Y Coordinate by finding the Height of an equilateral triangle of Side length 9 and extending that Height’s length up on Y Axis

Height = (9) * sqrt(3) * (1/2) = Y Coordinate

Thus, the coordinates of the 3rd vertex will be:

(4.5 ; 9 * sqrt(3) * (1/2) )

E

Posted from my mobile device

bumpbot · Answer

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In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9

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In the xy-plane, an equilateral triangle has vertices at (0, 0) and (9

Prep Toolkit

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