Question

in △abc, the coordinates of vertices a and b are a(-2,4) and b(-1,1). for each of the given coordinates of vertex c, is △abc a right triangle? select right triangle or not a right triangle for each set of coordinates.

Explanation:

Step1: Calculate the distance between two points

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(A(-2,4)\) and \(B(-1,1)\):
\(AB=\sqrt{(-1 + 2)^2+(1 - 4)^2}=\sqrt{1 + 9}=\sqrt{10}\)

Step2: For \(C(2,2)\)

\(AC=\sqrt{(2 + 2)^2+(2 - 4)^2}=\sqrt{16 + 4}=\sqrt{20}\)
\(BC=\sqrt{(2 + 1)^2+(2 - 1)^2}=\sqrt{9 + 1}=\sqrt{10}\)
Check \(AB^{2}+BC^{2}=10 + 10=20=AC^{2}\). So it is a right - triangle.

Step3: For \(C(0,4)\)

\(AC=\sqrt{(0 + 2)^2+(4 - 4)^2}=2\)
\(BC=\sqrt{(0 + 1)^2+(4 - 1)^2}=\sqrt{1+9}=\sqrt{10}\)
\(AB^{2}+AC^{2}=10 + 4 = 14
eq BC^{2}\), \(AB^{2}+BC^{2}=10+10 = 20
eq AC^{2}\), \(AC^{2}+BC^{2}=4 + 10=14
eq AB^{2}\). So it is not a right - triangle.

Step4: For \(C(-2,1)\)

\(AC=\sqrt{(-2 + 2)^2+(1 - 4)^2}=3\)
\(BC=\sqrt{(-2 + 1)^2+(1 - 1)^2}=1\)
\(AB^{2}=10\), \(AC^{2}+BC^{2}=9 + 1=10=AB^{2}\). So it is a right - triangle.

Answer:

For \(C(2,2)\): Right Triangle
For \(C(0,4)\): Not a Right Triangle
For \(C(-2,1)\): Right Triangle