Question

determine whether a triangle with the given vertices is a scalene, isosceles, or equilateral triangle. check all that apply.
(a) ( p(-8,3), q(-1,0), r(-1,6) )
(b) ( j(2,0), k(-1,-2), l(2,6) )
(c) ( d(3,1), e(-3,1), f(0,6) )

Explanation:

Step1: Calculate the side lengths using the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

• For part (a):
• \(PQ=\sqrt{(-1 + 8)^2+(0 - 3)^2}=\sqrt{49 + 9}=\sqrt{58}\)
• \(QR=\sqrt{(6 + 1)^2+(6 - 0)^2}=\sqrt{49+36}=\sqrt{85}\)
• \(PR=\sqrt{(6 + 8)^2+(6 - 3)^2}=\sqrt{196 + 9}=\sqrt{205}\)
• For part (b):
• \(JK=\sqrt{(-1 - 2)^2+(6 + 2)^2}=\sqrt{9 + 64}=\sqrt{73}\)
• \(KL=\sqrt{(2 + 1)^2+(6 - 6)^2}=\sqrt{9+0}=3\)
• \(JL=\sqrt{(2 - 2)^2+(6 + 2)^2}=\sqrt{0 + 64}=8\)
• For part (c):
• \(DE=\sqrt{(-3 - 3)^2+(6 - 6)^2}=\sqrt{36+0}=6\)
• \(EF=\sqrt{(0 + 3)^2+(6 - 6)^2}=\sqrt{9+0}=3\)
• \(DF=\sqrt{(0 - 3)^2+(6 - 6)^2}=\sqrt{9+0}=3\)

Step2: Classify the triangles

• For part (a): Since \(PQ

eq QR
eq PR\), it is a scalene triangle.

• For part (b): Since \(JK

eq KL
eq JL\), it is a scalene triangle.

• For part (c): Since \(EF = DF = 3\), it is an isosceles triangle.

Answer:

(a) Scalene triangle
(b) Scalene triangle
(c) Isosceles triangle