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5 votes
Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible coordinate of vertex R?

A. (-3, -6)
B. (0, 8)
C. (1, 8.9)
D. (1, -8.9)

User Duilio
by
7.4k points

2 Answers

1 vote

Answer:

1,8.9

Explanation:

User Shyamkkhadka
by
8.4k points
4 votes

Explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;


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


or \: \sqrt{( {5 + 3)}^(2) + ( {2 - 2)}^(2) }

By simplifying it we get,


8

Now,

again finding the distance between PR,


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

or,


\sqrt{( {x + 3)}^(2) + ( {y - 2)}^(2) }

By simplifying it we get,


= \sqrt{ {x}^(2) + {y}^(2) + 6x - 4y + 13 }

now, finding the distance of QR,


qr = \sqrt{( {x - 5)}^(2) + ( {y - 2)}^(2) }

or, by simplification we get,


\sqrt{ {x}^(2) + {y}^(2) - 10x - 4y + 29 }

now, equating PR and QR,


\sqrt{ {x}^(2) + {y}^(2) + 6x - 4y + 13} = \sqrt{ {x}^(2) + {y}^(2) - 10x - 4y + 29 }

we cancelled the root ,


{x}^(2) + {y}^(2) + 6x - 4y + 13 = {x}^(2) + {y}^(2) -10x - 4y + 29

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,


\sqrt{ {x}^(2) + {y}^(2) + 6x - 4y + 13 } = 8}

cancel the roots,


{x}^(2) + {y}^(2) + 6x - 4y + 13 = 8

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

Hope it helps...

User Yarning
by
8.3k points
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