Question

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?

Answer 1

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

$Distance = \sqrt{(x2-x1)^2+(y2-y1)^2}$

$PQ=\sqrt{(5-(-3))^2+(2-2)^2} = \sqrt{(5+3)^2+(0)^2} = \sqrt{8^2+0} = \sqrt{64} =8$

$QR=\sqrt{(x-5)^2+(y-2)^2} = \sqrt{x^2-10x+25 + y^2-4y+4} = \sqrt{x^2+y^2-10x-4y+29}$

$RS=\sqrt{(x-(-3)^2+(y-2)^2} = \sqrt{x^2+6x+9 + y^2-4y+4} = \sqrt{x^2+y^2+6x-4y+13}$.

We know, PQ= QR and PQ= RS.

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

Squaring both sides, we get

$x^2+y^2-10x-4y+29 =64$         -------- equation (1)

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

Squaring both sides, we get  

$x^2+y^2+6x-4y+13=64$            ---------- 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  

$(1)^2+y^2-10(1)-4y+29 = 64$

$1+y^2-10-4y+29 = 64$        (Simplifying)

$y^2-4y+20 = 64$

Subtracting 64 from both sides.

$y^2-4y+20-64 = 64-64$

$y^2-4y-44 = 0$

By using quadratic formula..

$y=\frac{-(-4)+\sqrt{(-4)^2-4(1)(-44)} }{2(1)} and$

$y=\frac{-(-4)-\sqrt{(-4)^2-4(1)(-44)} }{2(1)} [tex]y=2\left(1+2\sqrt{3}\right),\:y=2\left(1-2\sqrt{3}\right)$

In decimal form

y=8.928, -4.928.

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