Question

determine whether a triangle with the given vertices is a scalene, isosceles, or equilateral triangle. check all that apply
(a) ( e ( 1, 3 ), f ( 1, - 2 ) )
( - 3, 0 )
scalene triangle
isosceles triangle
equilateral triangle
(b) ( t ( 1, - 1 ), r ( 2, - 7 ), i ( 0, - 7 ) )
scalene triangle
isosceles triangle
equilateral triangle
(c) ( j ( 1, 0 ), k ( - 3, 1 ), l ( - 3, - 7 ) )
scalene triangle
isosceles triangle
equilateral triangle

Explanation:

Step1: Use 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): \(E(-3,0)\), \(F(1,-2)\), \(G(1,3)\)

• Calculate \(EF\):

\(EF=\sqrt{(1-(-3))^2+(-2 - 0)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\)

• Calculate \(FG\):

\(FG=\sqrt{(1 - 1)^2+(3-(-2))^2}=\sqrt{0 + 25}=5\)

• Calculate \(EG\):

\(EG=\sqrt{(1-(-3))^2+(3 - 0)^2}=\sqrt{16+9}=\sqrt{25}=5\)
Since \(FG = EG
eq EF\), it is an isosceles triangle.

For part (b): \(T(1,-1)\), \(R(2,-7)\), \(I(0,-7)\)

• Calculate \(TR\):

\(TR=\sqrt{(2 - 1)^2+(-7-(-1))^2}=\sqrt{1 + 36}=\sqrt{37}\)

• Calculate \(RI\):

\(RI=\sqrt{(0 - 2)^2+(-7-(-7))^2}=\sqrt{4+0}=2\)

• Calculate \(TI\):

\(TI=\sqrt{(0 - 1)^2+(-7-(-1))^2}=\sqrt{1 + 36}=\sqrt{37}\)
Since \(TR=TI
eq RI\), it is an isosceles triangle.

For part (c): \(J(1,0)\), \(K(-3,1)\), \(L(-3,-7)\)

• Calculate \(JK\):

\(JK=\sqrt{(-3 - 1)^2+(1 - 0)^2}=\sqrt{16 + 1}=\sqrt{17}\)

• Calculate \(KL\):

\(KL=\sqrt{(-3-(-3))^2+(-7 - 1)^2}=\sqrt{0+64}=8\)

• Calculate \(JL\):

\(JL=\sqrt{(-3 - 1)^2+(-7 - 0)^2}=\sqrt{16 + 49}=\sqrt{65}\)
Since \(JK
eq KL
eq JL\), it is a scalene triangle.

Answer:

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