QUESTION IMAGE
Question
determine whether a triangle with the given vertices is a right triangle.
| right triangle | not a right triangle | cannot be determined | |
|---|---|---|---|
| (b) ( t(-3, 1) ), ( r(1, 7) ), ( i(8, 0) ) | ( circ ) | ( circ ) | ( circ ) |
| (c) ( d(1, -2) ), ( e(-1, 4) ), ( f(5, 6) ) | ( circ ) | ( circ ) | ( circ ) |
To determine if a triangle with given vertices is a right triangle, we use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the lengths of all three sides, then check if the Pythagorean theorem (\( a^2 + b^2 = c^2 \), where \( c \) is the longest side) holds.
Part (a): Vertices \( P(4, 1) \), \( Q(2, -12) \), \( R(-2, -4) \)
Step 1: Calculate \( PQ \)
\( PQ = \sqrt{(2 - 4)^2 + (-12 - 1)^2} = \sqrt{(-2)^2 + (-13)^2} = \sqrt{4 + 169} = \sqrt{173} \)
Step 2: Calculate \( QR \)
\( QR = \sqrt{(-2 - 2)^2 + (-4 - (-12))^2} = \sqrt{(-4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80} \)
Step 3: Calculate \( PR \)
\( PR = \sqrt{(-2 - 4)^2 + (-4 - 1)^2} = \sqrt{(-6)^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \)
Step 4: Check Pythagorean theorem
Test \( (\sqrt{61})^2 + (\sqrt{80})^2 \stackrel{?}{=} (\sqrt{173})^2 \):
\( 61 + 80 = 141
eq 173 \).
Test other combinations (e.g., \( (\sqrt{61})^2 + (\sqrt{173})^2 \), \( (\sqrt{80})^2 + (\sqrt{173})^2 \))—none satisfy \( a^2 + b^2 = c^2 \).
Thus, \( PQR \) is not a right triangle.
Part (b): Vertices \( S(-3, 1) \), \( T(3, 1) \), \( U(8, 0) \)
Step 1: Calculate \( ST \)
\( ST = \sqrt{(3 - (-3))^2 + (1 - 1)^2} = \sqrt{(6)^2 + 0^2} = \sqrt{36} = 6 \)
Step 2: Calculate \( TU \)
\( TU = \sqrt{(8 - 3)^2 + (0 - 1)^2} = \sqrt{(5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \)
Step 3: Calculate \( SU \)
\( SU = \sqrt{(8 - (-3))^2 + (0 - 1)^2} = \sqrt{(11)^2 + (-1)^2} = \sqrt{121 + 1} = \sqrt{122} \)
Step 4: Check Pythagorean theorem
Test \( 6^2 + (\sqrt{26})^2 \stackrel{?}{=} (\sqrt{122})^2 \):
\( 36 + 26 = 62
eq 122 \).
Test other combinations—none satisfy \( a^2 + b^2 = c^2 \).
Thus, \( STU \) is not a right triangle.
Part (c): Vertices \( D(1, -2) \), \( E(-1, 4) \), \( F(5, 6) \)
Step 1: Calculate \( DE \)
\( DE = \sqrt{(-1 - 1)^2 + (4 - (-2))^2} = \sqrt{(-2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} \)
Step 2: Calculate \( EF \)
\( EF = \sqrt{(5 - (-1))^2 + (6 - 4)^2} = \sqrt{(6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} \)
Step 3: Calculate \( DF \)
\( DF = \sqrt{(5 - 1)^2 + (6 - (-2))^2} = \sqrt{(4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80} \)
Step 4: Check Pythagorean theorem
Test \( (\sqrt{40})^2 + (\sqrt{40})^2 \stackrel{?}{=} (\sqrt{80})^2 \):
\( 40 + 40 = 80 \), which matches \( (\sqrt{80})^2 = 80 \).
Thus, \( DEF \) is a right triangle (isosceles right triangle, since two legs are equal).
Final Answers:
- (a) \( PQR \): \(\boldsymbol{\text{Not a right triangle}}\)
- (b) \( STU \): \(\boldsymbol{\text{Not a right triangle}}\)
- (c) \( DEF \): \(\boldsymbol{\text{Right triangle}}\)
(Mark the appropriate columns in the table based on these results.)
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To determine if a triangle with given vertices is a right triangle, we use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the lengths of all three sides, then check if the Pythagorean theorem (\( a^2 + b^2 = c^2 \), where \( c \) is the longest side) holds.
Part (a): Vertices \( P(4, 1) \), \( Q(2, -12) \), \( R(-2, -4) \)
Step 1: Calculate \( PQ \)
\( PQ = \sqrt{(2 - 4)^2 + (-12 - 1)^2} = \sqrt{(-2)^2 + (-13)^2} = \sqrt{4 + 169} = \sqrt{173} \)
Step 2: Calculate \( QR \)
\( QR = \sqrt{(-2 - 2)^2 + (-4 - (-12))^2} = \sqrt{(-4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80} \)
Step 3: Calculate \( PR \)
\( PR = \sqrt{(-2 - 4)^2 + (-4 - 1)^2} = \sqrt{(-6)^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \)
Step 4: Check Pythagorean theorem
Test \( (\sqrt{61})^2 + (\sqrt{80})^2 \stackrel{?}{=} (\sqrt{173})^2 \):
\( 61 + 80 = 141
eq 173 \).
Test other combinations (e.g., \( (\sqrt{61})^2 + (\sqrt{173})^2 \), \( (\sqrt{80})^2 + (\sqrt{173})^2 \))—none satisfy \( a^2 + b^2 = c^2 \).
Thus, \( PQR \) is not a right triangle.
Part (b): Vertices \( S(-3, 1) \), \( T(3, 1) \), \( U(8, 0) \)
Step 1: Calculate \( ST \)
\( ST = \sqrt{(3 - (-3))^2 + (1 - 1)^2} = \sqrt{(6)^2 + 0^2} = \sqrt{36} = 6 \)
Step 2: Calculate \( TU \)
\( TU = \sqrt{(8 - 3)^2 + (0 - 1)^2} = \sqrt{(5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \)
Step 3: Calculate \( SU \)
\( SU = \sqrt{(8 - (-3))^2 + (0 - 1)^2} = \sqrt{(11)^2 + (-1)^2} = \sqrt{121 + 1} = \sqrt{122} \)
Step 4: Check Pythagorean theorem
Test \( 6^2 + (\sqrt{26})^2 \stackrel{?}{=} (\sqrt{122})^2 \):
\( 36 + 26 = 62
eq 122 \).
Test other combinations—none satisfy \( a^2 + b^2 = c^2 \).
Thus, \( STU \) is not a right triangle.
Part (c): Vertices \( D(1, -2) \), \( E(-1, 4) \), \( F(5, 6) \)
Step 1: Calculate \( DE \)
\( DE = \sqrt{(-1 - 1)^2 + (4 - (-2))^2} = \sqrt{(-2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} \)
Step 2: Calculate \( EF \)
\( EF = \sqrt{(5 - (-1))^2 + (6 - 4)^2} = \sqrt{(6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} \)
Step 3: Calculate \( DF \)
\( DF = \sqrt{(5 - 1)^2 + (6 - (-2))^2} = \sqrt{(4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80} \)
Step 4: Check Pythagorean theorem
Test \( (\sqrt{40})^2 + (\sqrt{40})^2 \stackrel{?}{=} (\sqrt{80})^2 \):
\( 40 + 40 = 80 \), which matches \( (\sqrt{80})^2 = 80 \).
Thus, \( DEF \) is a right triangle (isosceles right triangle, since two legs are equal).
Final Answers:
- (a) \( PQR \): \(\boldsymbol{\text{Not a right triangle}}\)
- (b) \( STU \): \(\boldsymbol{\text{Not a right triangle}}\)
- (c) \( DEF \): \(\boldsymbol{\text{Right triangle}}\)
(Mark the appropriate columns in the table based on these results.)