QUESTION IMAGE
Question
determine whether a triangle with the given vertices is a scalene, isosceles, or equilateral triangle. check all that apply.
(a) d(0, -3), e(5, 1), f(5, -7)
(b) p(1, -3), q(0, 2), r(-4, -3)
(c) a(-2, 3), b(-6, 4), c(0, 4)
scalene triangle
isosceles triangle
equilateral triangle
Part (a)
Step 1: Calculate \( DE \)
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) for \( D(0, -3) \) and \( E(5, 1) \):
\( DE = \sqrt{(5 - 0)^2 + (1 - (-3))^2} = \sqrt{25 + 16} = \sqrt{41} \)
Step 2: Calculate \( EF \)
For \( E(5, 1) \) and \( F(5, -7) \):
\( EF = \sqrt{(5 - 5)^2 + (-7 - 1)^2} = \sqrt{0 + 64} = 8 \)
Step 3: Calculate \( FD \)
For \( F(5, -7) \) and \( D(0, -3) \):
\( FD = \sqrt{(0 - 5)^2 + (-3 - (-7))^2} = \sqrt{25 + 16} = \sqrt{41} \)
Step 4: Classify the triangle
Since \( DE = FD = \sqrt{41} \) and \( EF = 8 \) (two sides equal), it is an isosceles triangle. It is not scalene (all sides different) or equilateral (all sides equal).
Part (b)
Step 1: Calculate \( PQ \)
For \( P(1, -3) \) and \( Q(0, 2) \):
\( PQ = \sqrt{(0 - 1)^2 + (2 - (-3))^2} = \sqrt{1 + 25} = \sqrt{26} \)
Step 2: Calculate \( QR \)
For \( Q(0, 2) \) and \( R(-4, -3) \):
\( QR = \sqrt{(-4 - 0)^2 + (-3 - 2)^2} = \sqrt{16 + 25} = \sqrt{41} \)
Step 3: Calculate \( RP \)
For \( R(-4, -3) \) and \( P(1, -3) \):
\( RP = \sqrt{(1 - (-4))^2 + (-3 - (-3))^2} = \sqrt{25 + 0} = 5 \)
Step 4: Classify the triangle
Since \( PQ = \sqrt{26} \), \( QR = \sqrt{41} \), \( RP = 5 \) (all sides different), it is a scalene triangle. It is not isosceles or equilateral.
Part (c)
Step 1: Calculate \( AB \)
For \( A(-2, 3) \) and \( B(-6, 4) \):
\( AB = \sqrt{(-6 - (-2))^2 + (4 - 3)^2} = \sqrt{16 + 1} = \sqrt{17} \)
Step 2: Calculate \( BC \)
For \( B(-6, 4) \) and \( C(0, 4) \):
\( BC = \sqrt{(0 - (-6))^2 + (4 - 4)^2} = \sqrt{36 + 0} = 6 \)
Step 3: Calculate \( CA \)
For \( C(0, 4) \) and \( A(-2, 3) \):
\( CA = \sqrt{(-2 - 0)^2 + (3 - 4)^2} = \sqrt{4 + 1} = \sqrt{5} \)
Step 4: Classify the triangle
Since \( AB = \sqrt{17} \), \( BC = 6 \), \( CA = \sqrt{5} \) (all sides different), it is a scalene triangle. It is not isosceles or equilateral.
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(a) Isosceles triangle (check "Isosceles triangle" for (a))
(b) Scalene triangle (check "Scalene triangle" for (b))
(c) Scalene triangle (check "Scalene triangle" for (c))