Question

determine whether a triangle with the given vertices is a right triangle.
(a) ( t(6,2), r(0,10), i(-5,-5) )
(b) ( j(-5,2), k(-1,6), l(9,-6) )
(c) ( p(2,6), q(-6,7), r(-8,-9) )
right triangle
not a right triangle
cannot be determined

Explanation:

Step1: Calculate the lengths of the sides using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

For part (a):

• \(TR=\sqrt{(0 - 6)^2+(10 - 2)^2}=\sqrt{36 + 64}=\sqrt{100} = 10\)
• \(TI=\sqrt{(-5 - 6)^2+(-5 - 2)^2}=\sqrt{121+49}=\sqrt{170}\)
• \(RI=\sqrt{(-5 - 0)^2+(-5 - 10)^2}=\sqrt{25 + 225}=\sqrt{250}\)

Check the Pythagorean theorem: \(10^2+(\sqrt{170})^2=100 + 170=270
eq250\)

For part (b):

• \(JK=\sqrt{(-1+5)^2+(6 - 2)^2}=\sqrt{16 + 16}=\sqrt{32}\)
• \(JL=\sqrt{(9 + 5)^2+(-6 - 2)^2}=\sqrt{196+64}=\sqrt{260}\)
• \(KL=\sqrt{(9 + 1)^2+(-6 - 6)^2}=\sqrt{100+144}=\sqrt{244}\)

Check the Pythagorean theorem: \((\sqrt{32})^2+(\sqrt{244})^2=32+244 = 276
eq260\)

For part (c):

• \(PQ=\sqrt{(-6 - 2)^2+(7 - 6)^2}=\sqrt{64 + 1}=\sqrt{65}\)
• \(PR=\sqrt{(-8 - 2)^2+(-9 - 6)^2}=\sqrt{100+225}=\sqrt{325}\)
• \(QR=\sqrt{(-8 + 6)^2+(-9 - 7)^2}=\sqrt{4+256}=\sqrt{260}\)

Check the Pythagorean theorem: \((\sqrt{65})^2+(\sqrt{260})^2=65+260=325\)

Answer:

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