Question

determine whether a triangle with the given vertices is a scalene, isosceles, or equilateral triangle. check all that apply.

	scalene triangle	isosceles triangle	equilateral triangle
(a) ( p(-8, 3), q(-1, 0), r(-1, 6) )	( square )	( square )	( square )
(b) ( d(3, 1), e(-3, 1), f(0, 6) )	( square )	( square )	( square )
(c) ( a(2, 0), b(-1, -2), c(2, 6) )	( square )	( square )	( square )

Explanation:

Part (a): Vertices \( P(-8, 3) \), \( Q(-1, 0) \), \( R(-1, 6) \)

Step 1: Calculate \( PQ \)

Use distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( PQ = \sqrt{(-1 - (-8))^2 + (0 - 3)^2} = \sqrt{(7)^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58} \)

Step 2: Calculate \( QR \)

\( QR = \sqrt{(-1 - (-1))^2 + (6 - 0)^2} = \sqrt{0 + 36} = 6 \)

Step 3: Calculate \( PR \)

\( PR = \sqrt{(-1 - (-8))^2 + (6 - 3)^2} = \sqrt{(7)^2 + (3)^2} = \sqrt{49 + 9} = \sqrt{58} \)
Since \( PQ = PR = \sqrt{58} \) and \( QR = 6 \), two sides are equal. So it is an isosceles triangle.

Part (b): Vertices \( D(3, 1) \), \( E(-3, 1) \), \( F(0, 6) \)

Step 1: Calculate \( DE \)

\( DE = \sqrt{(-3 - 3)^2 + (1 - 1)^2} = \sqrt{(-6)^2 + 0} = 6 \)

Step 2: Calculate \( EF \)

\( EF = \sqrt{(0 - (-3))^2 + (6 - 1)^2} = \sqrt{(3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \)

Step 3: Calculate \( DF \)

\( DF = \sqrt{(0 - 3)^2 + (6 - 1)^2} = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \)
Since \( EF = DF = \sqrt{34} \) and \( DE = 6 \), two sides are equal. So it is an isosceles triangle.

Part (c): Vertices \( A(2, 0) \), \( B(-1, -2) \), \( C(2, 6) \)

Step 1: Calculate \( AB \)

\( AB = \sqrt{(-1 - 2)^2 + (-2 - 0)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)

Step 2: Calculate \( BC \)

\( BC = \sqrt{(2 - (-1))^2 + (6 - (-2))^2} = \sqrt{(3)^2 + (8)^2} = \sqrt{9 + 64} = \sqrt{73} \)

Step 3: Calculate \( AC \)

\( AC = \sqrt{(2 - 2)^2 + (6 - 0)^2} = \sqrt{0 + 36} = 6 \)
Since all sides \( \sqrt{13} \), \( \sqrt{73} \), \( 6 \) are different, it is a scalene triangle.

Final Answers:

(a) Isosceles triangle (check the box for Isosceles triangle)
(b) Isosceles triangle (check the box for Isosceles triangle)
(c) Scalene triangle (check the box for Scalene triangle)

Answer:

Part (a): Vertices \( P(-8, 3) \), \( Q(-1, 0) \), \( R(-1, 6) \)

Step 1: Calculate \( PQ \)

Use distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( PQ = \sqrt{(-1 - (-8))^2 + (0 - 3)^2} = \sqrt{(7)^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58} \)

Step 2: Calculate \( QR \)

\( QR = \sqrt{(-1 - (-1))^2 + (6 - 0)^2} = \sqrt{0 + 36} = 6 \)

Step 3: Calculate \( PR \)

\( PR = \sqrt{(-1 - (-8))^2 + (6 - 3)^2} = \sqrt{(7)^2 + (3)^2} = \sqrt{49 + 9} = \sqrt{58} \)
Since \( PQ = PR = \sqrt{58} \) and \( QR = 6 \), two sides are equal. So it is an isosceles triangle.

Part (b): Vertices \( D(3, 1) \), \( E(-3, 1) \), \( F(0, 6) \)

Step 1: Calculate \( DE \)

\( DE = \sqrt{(-3 - 3)^2 + (1 - 1)^2} = \sqrt{(-6)^2 + 0} = 6 \)

Step 2: Calculate \( EF \)

\( EF = \sqrt{(0 - (-3))^2 + (6 - 1)^2} = \sqrt{(3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \)

Step 3: Calculate \( DF \)

\( DF = \sqrt{(0 - 3)^2 + (6 - 1)^2} = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \)
Since \( EF = DF = \sqrt{34} \) and \( DE = 6 \), two sides are equal. So it is an isosceles triangle.

Part (c): Vertices \( A(2, 0) \), \( B(-1, -2) \), \( C(2, 6) \)

Step 1: Calculate \( AB \)

\( AB = \sqrt{(-1 - 2)^2 + (-2 - 0)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)

Step 2: Calculate \( BC \)

\( BC = \sqrt{(2 - (-1))^2 + (6 - (-2))^2} = \sqrt{(3)^2 + (8)^2} = \sqrt{9 + 64} = \sqrt{73} \)

Step 3: Calculate \( AC \)

\( AC = \sqrt{(2 - 2)^2 + (6 - 0)^2} = \sqrt{0 + 36} = 6 \)
Since all sides \( \sqrt{13} \), \( \sqrt{73} \), \( 6 \) are different, it is a scalene triangle.

Final Answers:

(a) Isosceles triangle (check the box for Isosceles triangle)
(b) Isosceles triangle (check the box for Isosceles triangle)
(c) Scalene triangle (check the box for Scalene triangle)