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determine whether a triangle with the given vertices is a scalene, isos…

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(-1,0), q(7,3), r(7,-3) )
(b) ( a(0,1), b(-5,4), c(-5,-3) )
(c) ( j(6,5), k(-2,5), l(2,0) )

Explanation:

To determine the type of triangle (scalene, isosceles, equilateral) for each set of vertices, we use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the lengths of the sides.

Part (a): \( P(-1, 0) \), \( Q(7, 3) \), \( R(7, -3) \)
Step 1: Calculate \( PQ \)

\( PQ = \sqrt{(7 - (-1))^2 + (3 - 0)^2} = \sqrt{(8)^2 + (3)^2} = \sqrt{64 + 9} = \sqrt{73} \)

Step 2: Calculate \( QR \)

\( QR = \sqrt{(7 - 7)^2 + (-3 - 3)^2} = \sqrt{(0)^2 + (-6)^2} = \sqrt{0 + 36} = 6 \)

Step 3: Calculate \( PR \)

\( PR = \sqrt{(7 - (-1))^2 + (-3 - 0)^2} = \sqrt{(8)^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73} \)

Since \( PQ = PR = \sqrt{73} \) (two sides equal), the triangle is isosceles (not scalene or equilateral, since \( \sqrt{73}
eq 6 \)).

Part (b): \( A(0, 1) \), \( B(-5, 4) \), \( C(-5, -3) \)
Step 1: Calculate \( AB \)

\( AB = \sqrt{(-5 - 0)^2 + (4 - 1)^2} = \sqrt{(-5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34} \)

Step 2: Calculate \( BC \)

\( BC = \sqrt{(-5 - (-5))^2 + (-3 - 4)^2} = \sqrt{(0)^2 + (-7)^2} = \sqrt{0 + 49} = 7 \)

Step 3: Calculate \( AC \)

\( AC = \sqrt{(-5 - 0)^2 + (-3 - 1)^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \)

Since all sides (\( \sqrt{34} \), \( 7 \), \( \sqrt{41} \)) are distinct, the triangle is scalene (not isosceles or equilateral).

Part (c): \( J(6, 5) \), \( K(-2, 5) \), \( L(2, 0) \)
Step 1: Calculate \( JK \)

\( JK = \sqrt{(-2 - 6)^2 + (5 - 5)^2} = \sqrt{(-8)^2 + (0)^2} = \sqrt{64 + 0} = 8 \)

Step 2: Calculate \( KL \)

\( KL = \sqrt{(2 - (-2))^2 + (0 - 5)^2} = \sqrt{(4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \)

Step 3: Calculate \( JL \)

\( JL = \sqrt{(2 - 6)^2 + (0 - 5)^2} = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \)

Since \( KL = JL = \sqrt{41} \) (two sides equal), the triangle is isosceles (not scalene or equilateral, since \( 8
eq \sqrt{41} \)).

Final Answers:

(a) Isosceles triangle: \( \boldsymbol{\checkmark} \); Scalene/Equilateral: \( \boldsymbol{\square} \)
(b) Scalene triangle: \( \boldsymbol{\checkmark} \); Isosceles/Equilateral: \( \boldsymbol{\square} \)
(c) Isosceles triangle: \( \boldsymbol{\checkmark} \); Scalene/Equilateral: \( \boldsymbol{\square} \)

(Note: For each row, check the appropriate box. For (a), check "Isosceles triangle"; for (b), check "Scalene triangle"; for (c), check "Isosceles triangle".)

Answer:

To determine the type of triangle (scalene, isosceles, equilateral) for each set of vertices, we use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the lengths of the sides.

Part (a): \( P(-1, 0) \), \( Q(7, 3) \), \( R(7, -3) \)
Step 1: Calculate \( PQ \)

\( PQ = \sqrt{(7 - (-1))^2 + (3 - 0)^2} = \sqrt{(8)^2 + (3)^2} = \sqrt{64 + 9} = \sqrt{73} \)

Step 2: Calculate \( QR \)

\( QR = \sqrt{(7 - 7)^2 + (-3 - 3)^2} = \sqrt{(0)^2 + (-6)^2} = \sqrt{0 + 36} = 6 \)

Step 3: Calculate \( PR \)

\( PR = \sqrt{(7 - (-1))^2 + (-3 - 0)^2} = \sqrt{(8)^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73} \)

Since \( PQ = PR = \sqrt{73} \) (two sides equal), the triangle is isosceles (not scalene or equilateral, since \( \sqrt{73}
eq 6 \)).

Part (b): \( A(0, 1) \), \( B(-5, 4) \), \( C(-5, -3) \)
Step 1: Calculate \( AB \)

\( AB = \sqrt{(-5 - 0)^2 + (4 - 1)^2} = \sqrt{(-5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34} \)

Step 2: Calculate \( BC \)

\( BC = \sqrt{(-5 - (-5))^2 + (-3 - 4)^2} = \sqrt{(0)^2 + (-7)^2} = \sqrt{0 + 49} = 7 \)

Step 3: Calculate \( AC \)

\( AC = \sqrt{(-5 - 0)^2 + (-3 - 1)^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \)

Since all sides (\( \sqrt{34} \), \( 7 \), \( \sqrt{41} \)) are distinct, the triangle is scalene (not isosceles or equilateral).

Part (c): \( J(6, 5) \), \( K(-2, 5) \), \( L(2, 0) \)
Step 1: Calculate \( JK \)

\( JK = \sqrt{(-2 - 6)^2 + (5 - 5)^2} = \sqrt{(-8)^2 + (0)^2} = \sqrt{64 + 0} = 8 \)

Step 2: Calculate \( KL \)

\( KL = \sqrt{(2 - (-2))^2 + (0 - 5)^2} = \sqrt{(4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \)

Step 3: Calculate \( JL \)

\( JL = \sqrt{(2 - 6)^2 + (0 - 5)^2} = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \)

Since \( KL = JL = \sqrt{41} \) (two sides equal), the triangle is isosceles (not scalene or equilateral, since \( 8
eq \sqrt{41} \)).

Final Answers:

(a) Isosceles triangle: \( \boldsymbol{\checkmark} \); Scalene/Equilateral: \( \boldsymbol{\square} \)
(b) Scalene triangle: \( \boldsymbol{\checkmark} \); Isosceles/Equilateral: \( \boldsymbol{\square} \)
(c) Isosceles triangle: \( \boldsymbol{\checkmark} \); Scalene/Equilateral: \( \boldsymbol{\square} \)

(Note: For each row, check the appropriate box. For (a), check "Isosceles triangle"; for (b), check "Scalene triangle"; for (c), check "Isosceles triangle".)