Question

Which type of triangle is formed by joining the vertices A(-3, 6), B(2, 1), and C(9, 5)?

Answer 1

Answer:

An obtuse scalene triangle

Step-by-step explanation:

First, we calculate the length of the sides:

Side a – the distance between points B and C – is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d=\sqrt{(9-2)^2+(5-1)^2}\\d=\sqrt{(7)^2+(4)^2}\\d=\sqrt{49+16}\\d=\sqrt{65}\\d\approx{8.06}$

Side b – the distance between points C and A – is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d=\sqrt{(-3-9)^2+(6-5)^2}\\d=\sqrt{(-12)^2+(1)^2}\\d=\sqrt{144+1}\\d=\sqrt{145}\\d\approx{12.04}$

Side c – the distance between points A and B – is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d=\sqrt{(2-(-3))^2+(1-6)^2}\\d=\sqrt{(5)^2+(-5)^2}\\d=\sqrt{25+25}\\d=\sqrt{50}\\d\approx{7.07}$

Then, we calculate the measure of the angles:

$\angle{A}=\textrm{arccos}(\frac{b^2+c^2-a^2}{2bc})=\textrm{arccos}(\frac{12.04^2+7.07^2-8.06^2}{2(12.04)(7.07)})\approx{40.23^{\circ}}$

$\angle{B}=\textrm{arccos}(\frac{a^2+c^2-b^2}{2ac})=\textrm{arccos}(\frac{8.06^2+7.07^2-12.04^2}{2(8.06)(7.07)})\approx{105.27^{\circ}}$

$\angle{C}=180^{\circ}-40.23^{\circ}-105.27^{\circ}=34.50^{\circ}$

None of the sides are equal, so the triangle is not equilateral nor isosceles. None of the angles are 90°, so the triangle is not right. One of the angles is greater than 90°, so the triangle must be obtuse scalene.