Question

in △abc, the coordinates of vertices a and b are a(1, -1) and b(3, 2).
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.
right triangle not a right triangle
c(0, 2)
c(3, -1)
c(0, 4)

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 \(A(1,-1)\) and \(B(3,2)\):
\(AB=\sqrt{(3 - 1)^2+(2+ 1)^2}=\sqrt{4 + 9}=\sqrt{13}\)
For \(A(1,-1)\) and \(C(0,2)\):
\(AC=\sqrt{(0 - 1)^2+(2 + 1)^2}=\sqrt{1+9}=\sqrt{10}\)
For \(B(3,2)\) and \(C(0,2)\):
\(BC=\sqrt{(0 - 3)^2+(2 - 2)^2}=3\)
Check if \(AB^{2}+AC^{2}=BC^{2}\) or \(AB^{2}+BC^{2}=AC^{2}\) or \(AC^{2}+BC^{2}=AB^{2}\)
\(AB^{2}=13\), \(AC^{2}=10\), \(BC^{2}=9\)
\(10+9
eq13\), \(13 + 9
eq10\), \(10+13
eq9\)

For \(A(1,-1)\) and \(C(3,-1)\):
\(AC=\sqrt{(3 - 1)^2+(-1 + 1)^2}=2\)
For \(B(3,2)\) and \(C(3,-1)\):
\(BC=\sqrt{(3 - 3)^2+(-1 - 2)^2}=3\)
\(AB^{2}=13\), \(AC^{2}=4\), \(BC^{2}=9\)
\(4 + 9=13\)

For \(A(1,-1)\) and \(C(0,4)\):
\(AC=\sqrt{(0 - 1)^2+(4 + 1)^2}=\sqrt{1+25}=\sqrt{26}\)
For \(B(3,2)\) and \(C(0,4)\):
\(BC=\sqrt{(0 - 3)^2+(4 - 2)^2}=\sqrt{9+4}=\sqrt{13}\)
\(AB^{2}=13\), \(AC^{2}=26\), \(BC^{2}=13\)
\(13+13=26\)

Answer:

• \(C(0,2)\): Not a Right Triangle
• \(C(3,-1)\): Right Triangle
• \(C(0,4)\): Right Triangle