Question

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)

Answer 1

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..

Hope it helps...