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Nostrana [21]
3 years ago
15

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)
Mathematics
1 answer:
Law Incorporation [45]3 years ago
3 0

Step-by-step 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..

<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>

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