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-10 + 2t = -20 16t + 17t
2 answers:
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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
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-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).