Question

1. draw a scalene triangle on a coordinate plane, and use the distance formula to demonstrate that your triangle is scalene.

Explanation:

Step1: Choose three points

Let's choose \(A(0,0)\), \(B(3,0)\), \(C(1,2)\)

Step2: Calculate the distance between \(A\) and \(B\)

Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(A(0,0)\) and \(B(3,0)\):
\(d_{AB}=\sqrt{(3 - 0)^2+(0 - 0)^2}=\sqrt{9+0}=3\)

Step3: Calculate the distance between \(B\) and \(C\)

For \(B(3,0)\) and \(C(1,2)\):
\(d_{BC}=\sqrt{(1 - 3)^2+(2 - 0)^2}=\sqrt{(- 2)^2+2^2}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}\)

Step4: Calculate the distance between \(A\) and \(C\)

For \(A(0,0)\) and \(C(1,2)\):
\(d_{AC}=\sqrt{(1 - 0)^2+(2 - 0)^2}=\sqrt{1+4}=\sqrt{5}\)

Since \(3
eq2\sqrt{2}
eq\sqrt{5}\), the triangle with vertices \(A(0,0)\), \(B(3,0)\), \(C(1,2)\) is scalene.

Answer:

A scalene triangle with vertices \(A(0,0)\), \(B(3,0)\), \(C(1,2)\) is drawn. Using the distance formula, \(d_{AB} = 3\), \(d_{BC}=2\sqrt{2}\), \(d_{AC}=\sqrt{5}\). Since all three side - lengths are different, the triangle is scalene.