Question

the vertices of a triangle are located at $(-4,1),(0,-2)$ and $(3,2)$. select the correct option from the drop - down to complete the sentence
the triangle is

Explanation:

Step1: Recall distance formula

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Let the points be \(A(-4,1)\), \(B(0,-2)\), \(C(3,2)\).

Step2: Calculate \(AB\)

For \(A(-4,1)\) and \(B(0,-2)\):
\(AB=\sqrt{(0 - (-4))^2+(-2 - 1)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16 + 9}=\sqrt{25} = 5\)

Step3: Calculate \(BC\)

For \(B(0,-2)\) and \(C(3,2)\):
\(BC=\sqrt{(3 - 0)^2+(2 - (-2))^2}=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\)

Step4: Calculate \(AC\)

For \(A(-4,1)\) and \(C(3,2)\):
\(AC=\sqrt{(3 - (-4))^2+(2 - 1)^2}=\sqrt{(7)^2+(1)^2}=\sqrt{49 + 1}=\sqrt{50}=5\sqrt{2}\)

Step5: Check triangle type

We have \(AB = BC=5\), so two sides are equal. Now check if it's a right triangle. Using Pythagorean theorem: \(AB^2+BC^2=5^2 + 5^2=25 + 25 = 50\) and \(AC^2=(5\sqrt{2})^2 = 50\). So \(AB^2+BC^2=AC^2\), so it's a right triangle with \(AB = BC\), hence right isosceles.

Answer:

a right isosceles triangle