Respuesta :
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
[tex]Distance = \sqrt{(x2-x1)^2+(y2-y1)^2}[/tex]
[tex]PQ=\sqrt{(5-(-3))^2+(2-2)^2} = \sqrt{(5+3)^2+(0)^2} = \sqrt{8^2+0} = \sqrt{64} =8[/tex]
[tex]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}[/tex]
[tex]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}[/tex].
We know, PQ= QR and PQ= RS.
\sqrt{x^2+y^2-10x-4y+29}[/tex] = 8
Squaring both sides, we get
[tex]x^2+y^2-10x-4y+29 =64[/tex] -------- equation (1)
and \sqrt{x^2+y^2+6x-4y+13}[/tex] = 8
Squaring both sides, we get
[tex]x^2+y^2+6x-4y+13=64[/tex] ---------- 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
[tex](1)^2+y^2-10(1)-4y+29 = 64[/tex]
[tex]1+y^2-10-4y+29 = 64[/tex] (Simplifying)
[tex]y^2-4y+20 = 64[/tex]
Subtracting 64 from both sides.
[tex]y^2-4y+20-64 = 64-64[/tex]
[tex]y^2-4y-44 = 0[/tex]
By using quadratic formula..
[tex]y=\frac{-(-4)+\sqrt{(-4)^2-4(1)(-44)} }{2(1)} and[/tex]
[tex]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)[/tex]
In decimal form
y=8.928, -4.928.
Therefore, coordinates of vertex R could be (1, 8.928) or (1, -4.928).
The third coordinate of the equilateral triangle R is given as (1, 8.928) and (1, -4.928).
What is coordinate geometry?
Coordinate geometry is the study of geometry using the points in space. Using this, it is possible to find the distance between the points, the dividing line is m:n ratio, finding the mid-point of line, etc.
Carter drawers one side of the equilateral triangle ΔPQR on the coordinate plane at points P(-3, 2) and Q(5, 2). Then the third coordinate R will be
We know that an equilateral triangle has equal sides that are PQ = QR = RP.
[tex]\rm PQ = \sqrt{(5+3)^2+(2-2)^2}\\\\PQ = 8[/tex]
Let the coordinate of R be (x, y).
[tex]\rm QR = PQ\\\\(x-5)^2+(y-2)^2 = 8\\\\x^2 + y^2 - 10x-4y+21 = 0[/tex] ...1
And
[tex]\rm PR = PQ\\\\(x+3)^2+(y-2)^2 = 8\\\\x^2 + y^2 +6x-4y+5 = 0[/tex] ...2
On solving equations 1 and 2, we have
x = 1
For x = 1, the value of y will be 8.928 and -4.928.
More about the coordinate geometry link is given below.
https://brainly.com/question/1601567
