Question

try again
(a): your answer is incorrect.
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 )	( \bullet )	( circ )
(c) ( d(1, -2) ), ( e(-1, 4) ), ( f(5, 6) )	( \bullet )	( circ )	( circ )

Explanation:

Step1: Recall Distance Formula

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). We'll find the lengths of \(PQ\), \(QR\), and \(PR\) for part (a).

Step2: Calculate \(PQ\)

For \(P(4,11)\) and \(Q(2,-12)\):

$$
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$$

Step3: Calculate \(QR\)

For \(Q(2,-12)\) and \(R(-8,-4)\):

$$
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$$

Step4: Calculate \(PR\)

For \(P(4,11)\) and \(R(-8,-4)\):

$$
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$$

Step5: Check Pythagorean Theorem

Check if \(PQ^2=QR^2 + PR^2\), \(QR^2=PQ^2 + PR^2\), or \(PR^2=PQ^2 + QR^2\).

• \(PQ^2 = 533\), \(QR^2=164\), \(PR^2 = 369\)
• \(164+369 = 533\), so \(QR^2+PR^2 = PQ^2\). Thus, it is a right triangle.

Step6: Part (b) - Calculate Distances

For \(T(-3,1)\), \(R(1,7)\), \(I(8,0)\):

• \(TR\): \(\sqrt{(1 - (-3))^2+(7 - 1)^2}=\sqrt{16 + 36}=\sqrt{52}\)
• \(RI\): \(\sqrt{(8 - 1)^2+(0 - 7)^2}=\sqrt{49 + 49}=\sqrt{98}\)
• \(TI\): \(\sqrt{(8 - (-3))^2+(0 - 1)^2}=\sqrt{121 + 1}=\sqrt{122}\)
• Check Pythagorean: \(52 + 98 = 150

eq122\), \(52+122 = 174
eq98\), \(98 + 122 = 220
eq52\). Not a right triangle (correct as marked? Wait, no—wait, recalculate \(TI\): \((8 - (-3))=11\), \((0 - 1)=-1\), so \(11^2+(-1)^2 = 121 + 1 = 122\). \(TR^2=52\), \(RI^2=98\), \(TI^2=122\). \(52 + 98 = 150
eq122\), so not right (original mark was correct? Wait, no, maybe miscalculation. Wait, \(R(1,7)\), \(I(8,0)\): \(x\) difference 7, \(y\) difference -7, so \(7^2+(-7)^2 = 49 + 49 = 98\) (correct). \(T(-3,1)\) to \(R(1,7)\): \(x\) difference 4, \(y\) difference 6, \(16 + 36 = 52\) (correct). \(T(-3,1)\) to \(I(8,0)\): \(x\) difference 11, \(y\) difference -1, \(121 + 1 = 122\) (correct). So no right angle.

Step7: Part (c) - Calculate Distances

For \(D(1,-2)\), \(E(-1,4)\), \(F(5,6)\):

• \(DE\): \(\sqrt{(-1 - 1)^2+(4 - (-2))^2}=\sqrt{4 + 36}=\sqrt{40}\)
• \(EF\): \(\sqrt{(5 - (-1))^2+(6 - 4)^2}=\sqrt{36 + 4}=\sqrt{40}\)
• \(DF\): \(\sqrt{(5 - 1)^2+(6 - (-2))^2}=\sqrt{16 + 64}=\sqrt{80}\)
• Check Pythagorean: \((\sqrt{40})^2+(\sqrt{40})^2 = 40 + 40 = 80 = (\sqrt{80})^2\). So it is a right triangle (correct as marked).

But the error was in part (a): originally marked as not a right triangle, but it is. So for part (a), the correct option is "Right triangle".

Answer:

(a) Right triangle (the option "Right triangle" for part (a))
(b) Not a right triangle (as calculated, matches the mark)
(c) Right triangle (matches the mark)

But focusing on part (a) correction: the correct choice for (a) is "Right triangle".