Question

determine whether a triangle with the given vertices is a right triangle.

	right triangle	not a right triangle	cannot be determined
(b) ( p(6, 2) ), ( q(0, 10) ), ( r(-5, -5) )	( circ )	( circ )	( circ )
(c) ( a(2, 6) ), ( b(-6, 7) ), ( c(-8, -9) )	( circ )	( circ )	( circ )

Explanation:

Step1: Recall Distance Formula

The distance 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 a right triangle, the Pythagorean theorem \(a^2 + b^2 = c^2\) (where \(c\) is the longest side) must hold.

Part (a): Vertices \(T(-5, 2)\), \(R(-1, 6)\), \(I(9, -6)\)

Step2: Calculate \(TR\), \(RI\), \(TI\)

- \(TR\): \(\sqrt{(-1 - (-5))^2 + (6 - 2)^2} = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\)
- \(RI\): \(\sqrt{(9 - (-1))^2 + (-6 - 6)^2} = \sqrt{(10)^2 + (-12)^2} = \sqrt{100 + 144} = \sqrt{244} = 2\sqrt{61}\)
- \(TI\): \(\sqrt{(9 - (-5))^2 + (-6 - 2)^2} = \sqrt{(14)^2 + (-8)^2} = \sqrt{196 + 64} = \sqrt{260} = 2\sqrt{65}\)

Step3: Check Pythagorean Theorem

\((4\sqrt{2})^2 + (2\sqrt{61})^2 = 32 + 244 = 276\), \((2\sqrt{65})^2 = 260\). \(276 eq 260\). Also check other combinations: \((4\sqrt{2})^2 + (2\sqrt{65})^2 = 32 + 260 = 292 eq 244\); \((2\sqrt{61})^2 + (2\sqrt{65})^2 = 244 + 260 = 504 eq 32\). So not a right triangle? Wait, maybe I miscalculated. Wait, \(TR\): \((-1 - (-5)) = 4\), \(6 - 2 = 4\), so \(TR = \sqrt{16 + 16} = \sqrt{32}\). \(RI\): \(9 - (-1) = 10\), \(-6 - 6 = -12\), so \(RI = \sqrt{100 + 144} = \sqrt{244}\). \(TI\): \(9 - (-5) = 14\), \(-6 - 2 = -8\), so \(TI = \sqrt{196 + 64} = \sqrt{260}\). Wait, maybe slope method is better. Slope of \(TR\): \(\frac{6 - 2}{-1 - (-5)} = \frac{4}{4} = 1\). Slope of \(RI\): \(\frac{-6 - 6}{9 - (-1)} = \frac{-12}{10} = -\frac{6}{5}\). Slope of \(TI\): \(\frac{-6 - 2}{9 - (-5)} = \frac{-8}{14} = -\frac{4}{7}\). No slopes are negative reciprocals, so not perpendicular. Wait, but maybe I made a mistake. Wait, let's recalculate distances:

Wait, \(TR\): \(\sqrt{(-1 +5)^2 + (6 - 2)^2} = \sqrt{16 + 16} = \sqrt{32}\). \(RI\): \(\sqrt{(9 +1)^2 + (-6 -6)^2} = \sqrt{100 + 144} = \sqrt{244}\). \(TI\): \(\sqrt{(9 +5)^2 + (-6 -2)^2} = \sqrt{196 + 64} = \sqrt{260}\). Now, check \(TR^2 + RI^2 = 32 + 244 = 276\), \(TI^2 = 260\). \(TR^2 + TI^2 = 32 + 260 = 292\), \(RI^2 = 244\). \(RI^2 + TI^2 = 244 + 260 = 504\), \(TR^2 = 32\). So no, not a right triangle? Wait, but maybe I messed up. Wait, let's do part (a) again. Wait, maybe the correct answer is right triangle? Wait, no, let's check slopes. Slope of \(TR\): 1, slope of \(TI\): \(-4/7\), not negative reciprocal. Slope of \(TR\) and \(RI\): 1 and -6/5, not negative reciprocal. Slope of \(RI\) and \(TI\): -6/5 and -4/7, not negative reciprocal. So part (a) is not a right triangle? Wait, maybe I made a mistake. Let's check part (b): \(P(6,2)\), \(Q(0,10)\), \(R(-5,-5)\).

Part (b): Vertices \(P(6, 2)\), \(Q(0, 10)\), \(R(-5, -5)\)

Step2: Calculate distances

- \(PQ\): \(\sqrt{(0 - 6)^2 + (10 - 2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)
- \(QR\): \(\sqrt{(-5 - 0)^2 + (-5 - 10)^2} = \sqrt{25 + 225} = \sqrt{250} = 5\sqrt{10}\)
- \(PR\): \(\sqrt{(-5 - 6)^2 + (-5 - 2)^2} = \sqrt{121 + 49} = \sqrt{170}\)

Step3: Check Pythagorean Theorem

\(PQ^2 + PR^2 = 100 + 170 = 270\), \(QR^2 = 250\). \(PQ^2 + QR^2 = 100 + 250 = 350\), \(PR^2 = 170\). \(QR^2 + PR^2 = 250 + 170 = 420\), \(PQ^2 = 100\). No. Wait, slope method: slope of \(PQ\): \(\frac{10 - 2}{0 - 6} = \frac{8}{-6} = -\frac{4}{3}\). Slope of \(QR\): \(\frac{-5 - 10}{-5 - 0} = \frac{-15}{-5} = 3\). Slope of \(PR\): \(\frac{-5 - 2}{-5 - 6} = \frac{-7}{-11} = \frac{7}{11}\). Now, slope of \(PQ\) is \(-4/3\), slope of \(QR\) is 3. The product of slopes: \(-4/3 * 3 = -4 eq -1\). Wait, no, 3 is the negative reciprocal of -1/3, not -4/3. Wait, \(-4/3\) and \(3/4\) would be negative reciprocal. Wait, maybe I miscalculated \(PQ\) distance. \(PQ\): \(0 - 6 = -6\), \(10 - 2 = 8\), so distance is \(\sqrt{36 + 64} = \sqrt{100} = 10\). \(QR\): \(-5 - 0 = -5\), \(-5 - 10 = -15\), distance \(\sqrt{25 + 225} = \sqrt{250}\). \(PR\): \(-5 - 6 = -11\), \(-5 - 2 = -7\), distance \(\sqrt{121 + 49} = \sqrt{170}\). Wait, maybe part (b) is a right triangle? Wait, no, let's check part (c): \(A(2,6)\), \(B(-6,7)\), \(C(-8,-9)\).

Part (c): Vertices \(A(2, 6)\), \(B(-6, 7)\), \(C(-8, -9)\)

Step2: Calculate distances

- \(AB\): \(\sqrt{(-6 - 2)^2 + (7 - 6)^2} = \sqrt{64 + 1} = \sqrt{65}\)
- \(BC\): \(\sqrt{(-8 - (-6))^2 + (-9 - 7)^2} = \sqrt{4 + 256} = \sqrt{260}\)
- \(AC\): \(\sqrt{(-8 - 2)^2 + (-9 - 6)^2} = \sqrt{100 + 225} = \sqrt{325}\)

Step3: Check Pythagorean Theorem

\(AB^2 + BC^2 = 65 + 260 = 325\), which equals \(AC^2 = 325\). So \(AB^2 + BC^2 = AC^2\), so it is a right triangle.

Wait, let's recheck part (c):

\(AB\): \(x\)-difference: \(-6 - 2 = -8\), \(y\)-difference: \(7 - 6 = 1\), so \(AB = \sqrt{(-8)^2 + 1^2} = \sqrt{64 + 1} = \sqrt{65}\).

\(BC\): \(x\)-difference: \(-8 - (-6) = -2\), \(y\)-difference: \(-9 - 7 = -16\), so \(BC = \sqrt{(-2)^2 + (-16)^2} = \sqrt{4 + 256} = \sqrt{260}\).

\(AC\): \(x\)-difference: \(-8 - 2 = -10\), \(y\)-difference: \(-9 - 6 = -15\), so \(AC = \sqrt{(-10)^2 + (-15)^2} = \sqrt{100 + 225} = \sqrt{325}\).

Now, \(AB^2 = 65\), \(BC^2 = 260\), \(AC^2 = 325\). \(65 + 260 = 325\), so yes, right triangle.

For part (a):

\(TR\): \(\sqrt{(-1 +5)^2 + (6 - 2)^2} = \sqrt{16 + 16} = \sqrt{32}\)

\(RI\): \(\sqrt{(9 +1)^2 + (-6 -6)^2} = \sqrt{100 + 144} = \sqrt{244}\)

\(TI\): \(\sqrt{(9 +5)^2 + (-6 -2)^2} = \sqrt{196 + 64} = \sqrt{260}\)

Check \(TR^2 + TI^2 = 32 + 260 = 292\), \(RI^2 = 244\). \(TR^2 + RI^2 = 32 + 244 = 276\), \(TI^2 = 260\). \(RI^2 + TI^2 = 244 + 260 = 504\), \(TR^2 = 32\). No.

Part (b):

\(PQ\): \(\sqrt{(0 - 6)^2 + (10 - 2)^2} = \sqrt{36 + 64} = 10\)

\(QR\): \(\sqrt{(-5 - 0)^2 + (-5 - 10)^2} = \sqrt{25 + 225} = \sqrt{250}\)

\(PR\): \(\sqrt{(-5 - 6)^2 + (-5 - 2)^2} = \sqrt{121 + 49} = \sqrt{170}\)

Check \(10^2 + (\sqrt{170})^2 = 100 + 170 = 270\), \((\sqrt{250})^2 = 250\). \(10^2 + (\sqrt{250})^2 = 100 + 250 = 350\), \((\sqrt{170})^2 = 170\). \((\sqrt{250})^2 + (\sqrt{170})^2 = 250 + 170 = 420\), \(10^2 = 100\). No.

So:

(a) Not a right triangle? Wait, but maybe I made a mistake. Wait, let's use slope for part (a):

Slope of \(TR\): (6-2)/(-1 - (-5)) = 4/4 = 1

Slope of \(TI\): (-6 - 2)/(9 - (-5)) = -8/14 = -4/7

Slope of \(RI\): (-6 - 6)/(9 - (-1)) = -12/10 = -6/5

No two slopes are negative reciprocals, so no right angle.

(b) Slope of \(PQ\): (10-2)/(0-6) = 8/-6 = -4/3

Slope of \(QR\): (-5 -10)/(-5 - 0) = -15/-5 = 3

Slope of \(PR\): (-5 -2)/(-5 -6) = -7/-11 = 7/11

Product of \(PQ\) and \(QR\) slopes: (-4/3)*3 = -4 ≠ -1. So not perpendicular.

(c) Slope of \(AB\): (7-6)/(-6 -2) = 1/-8 = -1/8

Slope of \(BC\): (-9 -7)/(-8 - (-6)) = -16/-2 = 8

Product of \(AB\) and \(BC\) slopes: (-1/8)*8 = -1. Oh! Wait, I made a mistake earlier. Slope of \(AB\) is (7-6)/(-6 -2) = 1/(-8) = -1/8. Slope of \(BC\) is (-9 -7)/(-8 - (-6)) = -16/(-2) = 8. So (-1/8)*8 = -1, so they are perpendicular. So \(AB \perp BC\), so right triangle. And distance check: \(AB^2 + BC^2 = 65 + 260 = 325 = AC^2\), which matches. So part (c) is right triangle.

So:

(a) Let's recheck slopes. Slope of \(TR\): 1, slope of \(RI\): -6/5, slope of \(TI\): -4/7. No negative reciprocals. So not right triangle.

(b) Slope of \(PQ\): -4/3, slope of \(QR\): 3, product -4 ≠ -1. Slope of \(PQ\) and \(PR\): -4/3 and 7/11, product -28/33 ≠ -1. Slope of \(QR\) and \(PR\): 3 and 7/11, product 21/11 ≠ -1. So not right triangle.

(c) Slope of \(AB\) and \(BC\) are negative reciprocals, so right triangle.

Answer:

(a) Not a right triangle
(b) Not a right triangle
(c) Right triangle

(For each part, mark the corresponding option: (a) "Not a right triangle", (b) "Not a right triangle", (c) "Right triangle")