Carter drawers one side of equilateral triangle pqr on the coordinate plane at points p(-3, 2) and q(5, 2). which ordered pair is a possible coordinate of vertex r?

Given: Two coordinates of an equilateral triangle P(-3, 2) and Q(5, 2).

To find: Coordinate of third vertex R.

Solution: Let us take coordinate of R is (x,y).

Because we are given PQR is an equilateral triangle, all side of the triangle PQR are equal.

We can say PQ = QR = RS.

Let us find equations by using distance formula.

Distance between two coordinates is given by formula

.

We know, PQ= QR and PQ= RS.

\sqrt{x^2+y^2-10x-4y+29}[/tex] = 8

Squaring both sides, we get

        -------- equation (1)

and \sqrt{x^2+y^2+6x-4y+13}[/tex] = 8

Squaring both sides, we get  

           ---------- equation (2).

Subtracting equation 2 from equation 1.

[/tex]x^2+y^2-10x-4y+29[/tex] = 64

[/tex]x^2+y^2+6x-4y+13[/tex] = 64

----------------------------------------------------

-16x +16 = 0

Subtracting 16 on both sides ,

-16x +16-16 = 0-16

-16x = -16

Dividing -16 on both sides.

-16x/-16 = -16/-16

x=1.

Plugging x=1 in first equation  

       (Simplifying)

Subtracting 64 from both sides.

By using quadratic formula..

In decimal form

y=8.928, -4.928.

Therefore, coordinates of vertex R could be (1, 8.928) or (1, -4.928).

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

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