Question

classify \\( \triangle a b c \\) by its sides.
then determine whether it is
a right triangle.
\\( a ( 3,3 ), b ( 6,9 ), c ( 6, - 3 ) \\)
the triangle can be classified by
its sides as

this triangle a right triangle.

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 side \(AB\) with \(A(3,3)\) and \(B(6,9)\):
\(AB=\sqrt{(6 - 3)^2+(9 - 3)^2}=\sqrt{3^2+6^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}\)
For side \(BC\) with \(B(6,9)\) and \(C(6,-3)\):
\(BC=\sqrt{(6 - 6)^2+(-3 - 9)^2}=\sqrt{0+(- 12)^2}=\sqrt{144}=12\)
For side \(AC\) with \(A(3,3)\) and \(C(6,-3)\):
\(AC=\sqrt{(6 - 3)^2+(-3 - 3)^2}=\sqrt{3^2+(-6)^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}\)

Step2: Classify the triangle by its sides

Since \(AB = AC=3\sqrt{5}\) and \(BC = 12\), the triangle is isosceles (two sides are equal).

Step3: Check if it is a right - triangle using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c\) is the longest side)

Let \(a = 3\sqrt{5}\), \(b = 3\sqrt{5}\), \(c = 12\)
\(a^{2}+b^{2}=(3\sqrt{5})^{2}+(3\sqrt{5})^{2}=45 + 45=90\)
\(c^{2}=12^{2}=144\)
Since \(90
eq144\), it is not a right - triangle.

Answer:

The triangle can be classified by its sides as isosceles. This triangle is not a right triangle.