Question

what is the perimeter of ^PQR with vertices P(-2,9) Q(7,-3) and R(-2,-3) in the coordinate plane?

Answer 1

Using the distance formula we can determine the perimeter of PQR.

Distance formula: d=√(x-x1)^2+(y-y1)^2
distance of PR: √(-2+2)^2+(9+3)^2= 12
distance of QR: √(7+2)^2+(-3+3)^2=9
distance of PQ: √(7+2)^2+(-3-9)^2= 15

Perimeter: 12+9+15=36

Answer 2

The perimeter of the given triangle in the coordinate plane is 36 units.

Given:

The triangle PQR with vertices P(-2,9) Q(7,-3) and R(-2,-3) in the coordinate plane.

To find:

The perimeter of the triangle.

Solution:

The distance formula to find the distance between two points ($x_1,y_1$) and

($x_2,y_2$):

$d=\sqr{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between point  P(-2,9) and Q(7,-3)

$PQ=\sqrt{(7-(-2))^2+((-3)-9)^2}\\\\PQ=\sqrt{(9)^2+(-12)^2}\\\\PQ=15 units$

The distance between point Q(7,-3) and R(-2,-3)

$QR=\sqrt{((-2)-7)^2+((-3)-(-3))^2}\\\\QR=\sqrt{(-9)^2+(0)^2}\\\\QR=9units$

The distance between point R(-2,-3) and  P(-2,9)

$RP =\sqrt{((-2)-(-2))^2+(9-(-3))^2}\\\\RP =\sqrt{(0)^2+(12)^2}\\\\RP =12units$

The perimeter of the triangle will be given by"

$P= PQ+QR+RP\\\\=15 units+9 units+12 units = 36 units$

The perimeter of the given triangle in the coordinate plane is 36 units.

Learn more about the distance formula here:

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