Question

classifying triangles (in the coordinate plane) classify: 21. 22. 23. find the measures of the sides of \\( \triangle jkl \\), then classify it by its sides. 21. \\( j(-7, -7) \\), \\( k(-9, 1) \\), \\( l(-1, -1) \\) 22. \\( j(-3, 2) \\), \\( k(2, 6) \\), \\( l(8, -1) \\) 23. \\( j(1, -13) \\), \\( k(3, 3) \\), \\( l(10, -6) \\) \\( \copyright \\) gina wilson (all things algebra\\( \circledr \\), llc), 2014-2020

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 apply this to find the lengths of \(JK\), \(KL\), and \(JL\) (or corresponding sides for each triangle) to classify the triangle by its sides.

For Problem 21: \(J(-7, -7)\), \(K(-9, 1)\), \(L(-1, -1)\)

Step2: Calculate \(JK\)

\(x_1=-7,y_1=-7; x_2=-9,y_2=1\)
\(JK = \sqrt{(-9 - (-7))^2 + (1 - (-7))^2} = \sqrt{(-2)^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}\)

Step3: Calculate \(KL\)

\(x_1=-9,y_1=1; x_2=-1,y_2=-1\)
\(KL = \sqrt{(-1 - (-9))^2 + (-1 - 1)^2} = \sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}\)

Step4: Calculate \(JL\)

\(x_1=-7,y_1=-7; x_2=-1,y_2=-1\)
\(JL = \sqrt{(-1 - (-7))^2 + (-1 - (-7))^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\)

Since \(JK = KL\), \(\triangle JKL\) is isosceles.

For Problem 22: \(J(-3, 2)\), \(K(2, 6)\), \(L(8, -1)\)

Step2: Calculate \(JK\)

\(x_1=-3,y_1=2; x_2=2,y_2=6\)
\(JK = \sqrt{(2 - (-3))^2 + (6 - 2)^2} = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}\)

Step3: Calculate \(KL\)

\(x_1=2,y_1=6; x_2=8,y_2=-1\)
\(KL = \sqrt{(8 - 2)^2 + (-1 - 6)^2} = \sqrt{6^2 + (-7)^2} = \sqrt{36 + 49} = \sqrt{85}\)

Step4: Calculate \(JL\)

\(x_1=-3,y_1=2; x_2=8,y_2=-1\)
\(JL = \sqrt{(8 - (-3))^2 + (-1 - 2)^2} = \sqrt{11^2 + (-3)^2} = \sqrt{121 + 9} = \sqrt{130}\)

Since \(JK
eq KL
eq JL\), \(\triangle JKL\) is scalene.

For Problem 23: \(J(1, -13)\), \(K(3, 3)\), \(L(10, -6)\)

Step2: Calculate \(JK\)

\(x_1=1,y_1=-13; x_2=3,y_2=3\)
\(JK = \sqrt{(3 - 1)^2 + (3 - (-13))^2} = \sqrt{2^2 + 16^2} = \sqrt{4 + 256} = \sqrt{260} = 2\sqrt{65}\)

Step3: Calculate \(KL\)

\(x_1=3,y_1=3; x_2=10,y_2=-6\)
\(KL = \sqrt{(10 - 3)^2 + (-6 - 3)^2} = \sqrt{7^2 + (-9)^2} = \sqrt{49 + 81} = \sqrt{130}\)

Step4: Calculate \(JL\)

\(x_1=1,y_1=-13; x_2=10,y_2=-6\)
\(JL = \sqrt{(10 - 1)^2 + (-6 - (-13))^2} = \sqrt{9^2 + 7^2} = \sqrt{81 + 49} = \sqrt{130}\)

Since \(KL = JL\), \(\triangle JKL\) is isosceles.

Answer:

• Problem 21: Isosceles Triangle (two sides \(2\sqrt{17}\), one side \(6\sqrt{2}\))
• Problem 22: Scalene Triangle (all sides \(\sqrt{41}\), \(\sqrt{85}\), \(\sqrt{130}\) are distinct)
• Problem 23: Isosceles Triangle (two sides \(\sqrt{130}\), one side \(2\sqrt{65}\))