triangle PQR has vertices p(-3,5),q(7,3) and r(-1,-5) find the distance between PQ,QR and RP

simply use the "length of line between 2 points" formula.

I will do PQ, you do the other two

PQ = √( (7+3)^2 + (3-5)^2 )
= √(100 + 4)
= √104 = √4 √26
= 2√26

To find the distance between two points in a coordinate plane, we can use the distance formula:

The formula for the distance between two points (x1, y1) and (x2, y2) is given by:

Distance = √((x2 - x1)^2 + (y2 - y1)^2)

Let's calculate the distances between the given points:

1. Distance between points P(-3, 5) and Q(7, 3) (PQ):
Distance(PQ) = √((7 - (-3))^2 + (3 - 5)^2)
= √((7 + 3)^2 + (-2)^2)
= √(10^2 + 4)
= √(100 + 4)
= √104
≈ 10.2 (rounded to one decimal place)

2. Distance between points Q(7, 3) and R(-1, -5) (QR):
Distance(QR) = √((-1 - 7)^2 + (-5 - 3)^2)
= √((-8)^2 + (-8)^2)
= √(64 + 64)
= √128
≈ 11.3 (rounded to one decimal place)

3. Distance between points R(-1, -5) and P(-3, 5) (RP):
Distance(RP) = √((-3 - (-1))^2 + (5 - (-5))^2)
= √((-3 + 1)^2 + (5 + 5)^2)
= √((-2)^2 + 10^2)
= √(4 + 100)
= √104
≈ 10.2 (rounded to one decimal place)

Therefore, the distances between the points PQ, QR, and RP are approximately 10.2, 11.3, and 10.2 units, respectively.