Find the area and perimeter of the figure determined by the points. (−7, 2)​, (−3, 1)​, (−2, 5)

Question

Find the area and perimeter of the figure determined by the points. (−7, 2)​, (−3, 1)​, (−2, 5).

Answer 1

(-7, 2), (-3, 1), (-2,5)

are vertices of a theoretical triangle

whose sides have lines with slope

= 1/4, -4, and 3/5

It's a right triangle since 1/4 and -4 are negative inverses

area = 1/2 base times height

height = sqr17

base = sqr17

1/2 x sqr17 x sqr17 = 17/2

Perimeter = base + height + hypotenuse = sqr17+sqr17+sqr34= 2sqr17 +sqr34

= (2+sqr2)sqr17

or use heron's formula for the area

s = the semiperimeter = 17/4

Area = sqr(s(s-a)(s-b)(s-c)) where a, b and c are the lengths of each of the 3 sides

= sqr(17/4)(17/4 - sqr17)(17/4 -sqr17)(17/4- sqr34)

=

Answer 2

isn't the area =17/2, with sides of √17, √17 √ 34

sides are calculated with distance formula.

by observation, side1²+side2²=side3²

triangle is a right triangle with legs of √17, and hypotenuse √34

area of right triangle is (1/2)(leg side1)(leg side2)

Answer 3

You find the perimeter by adding the distances between the points. I'll do the first side (let's call it a)

a = sqrt((-7-(-3))² + (2-1)²) = sqrt(17)

find b and c using the other pairs of points

Perimeter = a + b + c

area = sqrt(s(s-a)(s-b)(s-c)) where s is P/2 (Heron's formula - one way to do it)

Please consider a tutor.