Question

the vertices of a triangle are listed below.
which of the following correctly classifies the triangle?
a. the triangle is a right scalene triangle.
b. the triangle is a right isosceles triangle.
c. the triangle is an acute scalene triangle.
d. the triangle is an acute isosceles triangle.
a(-5,6), b(-13,0), c(-5,-6)

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 \(AB\):
\(x_1=-5,y_1 = 6,x_2=-13,y_2 = 0\)
\(AB=\sqrt{(-13 + 5)^2+(0 - 6)^2}=\sqrt{(-8)^2+(-6)^2}=\sqrt{64 + 36}=\sqrt{100}=10\)
For \(AC\):
\(x_1=-5,y_1 = 6,x_2=-5,y_2=-6\)
\(AC=\sqrt{(-5+5)^2+(-6 - 6)^2}=\sqrt{0+(-12)^2}=12\)
For \(BC\):
\(x_1=-13,y_1 = 0,x_2=-5,y_2=-6\)
\(BC=\sqrt{(-5 + 13)^2+(-6-0)^2}=\sqrt{8^2+(-6)^2}=\sqrt{64 + 36}=\sqrt{100}=10\)

Step2: Check the type of triangle

Since \(AB = BC=10\), two sides are equal.
Also, check if it is a right - triangle using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Let \(a = 10,b = 10,c = 12\)
\(a^{2}+b^{2}=10^{2}+10^{2}=200\) and \(c^{2}=12^{2}=144\), \(a^{2}+b^{2}
eq c^{2}\)
Since two sides are equal (\(AB = BC\)) and all angles are acute (using the fact that \(a^{2}+b^{2}>c^{2}\) for all combinations of sides in an acute triangle, \(10^{2}+10^{2}>12^{2},10^{2}+12^{2}>10^{2}\))

Answer:

D. The triangle is an acute isosceles triangle.