Arectangle is a four sided flat shape where every integer angle is a right angle. therefore, opposite sides are parallel and consecutive sides are perpendicular. rae is drawing rectangle pqrs on a coordinate plane. the rectangle has coordinates p (-1,2), q(2,4), r(x, y) and s(3,-4)

we know that

The rectangle has coordinates P (-1,2), Q(2,4), R(x,y) and S(3,-4)

using a graph tool

see the attached figure

Step 1

Find the slope of the line PS

the slope is equal to

m=(y2-y1)/(x2-x1)

mPS=(-4-2)/(3+1)=-6/4=-3/2

Step 2

Find the equation of the line QR

we know that

PS and QR are parallel lines

so

their slopes are the same

mQR=-3/2

with mQR and the point Q(2,4) find the equation of the line

y-y1=m*(x-x1)

y-4=(-3/2)*(x-2)

y=(-3/2)*x+3+4--------> y=7-(3/2)*x

Step 3

Find the equation of the line SR

we know that

SR and QR are perpendicular lines

so

the product of their slopes is equal to minus one

mSR*mQR=-1

mSR=2/3

with mSR and the point S(3,-4) find the equation of the line SR

y-y1=m*(x-x1)

y+4=(2/3)*(x-3)

y=(2/3)*x-2-4---------> y=-6+(2/3)*x

Step 4

Find the intersection of the lines QR and SR

we know that

the intersection of the lines QR and SR is the point R

y=7-(3/2)*x------> equation 1

y=-6+(2/3)*x------> equation 2

equate equation 1 and equation 2

7-(3/2)*x=-6+(2/3)*x

7+6=(2/3)*x+(3/2)*x

13=(2/3)*x+(3/2)*x

Multiply by 6 both sides

6*13=4*x+9*x

13*x=6*13

x=6

find the value of y

y=-6+(2/3)*x------> y=-6+(2/3)*6---------> y=-6+4

y=-2

therefore

the answer is

The coordinates of point R are (6,-2)

see the attached figure N 2

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step-by-step explanation: