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.C(0,2) C(3,−1) C(0,4)

we know thatthe formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
In this problem we have[tex]A(1,-1)\ B(3,2)\ C1(0,2)\ C2(3,-1)\ C3(0,4)[/tex]Step 1Find the distance AB[tex]A(1,-1)\ B(3,2)[/tex]Substitute the values in the formula[tex]d=\sqrt{(2+1)^{2}+(3-1)^{2}}[/tex]
[tex]d=\sqrt{(3)^{2}+(2)^{2}}[/tex]
[tex]dAB=\sqrt{13}\ units[/tex]
Step 2Find the distance AC1[tex]A(1,-1)\ C1(0,2)[/tex]Substitute the values in the formula[tex]d=\sqrt{(2+1)^{2}+(0-1)^{2}}[/tex]
[tex]d=\sqrt{(3)^{2}+(-1)^{2}}[/tex]
[tex]dAC1=\sqrt{10}\ units[/tex]
Step 3Find the distance BC1[tex]B(3,2)\ C1(0,2)[/tex]Substitute the values in the formula[tex]d=\sqrt{(2-2)^{2}+(0-3)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(-3)^{2}}[/tex]
[tex]dBC1=3\ units[/tex]
Step 4Find the distance AC2[tex]A(1,-1)\ C2(3,-1)[/tex]Substitute the values in the formula[tex]d=\sqrt{(-1+1)^{2}+(3-1)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(2)^{2}}[/tex]
[tex]dAC2=2\ units[/tex]
Step 5Find the distance BC2[tex]B(3,2)\ C2(3,-1)[/tex]Substitute the values in the formula[tex]d=\sqrt{(-1-2)^{2}+(3-3)^{2}}[/tex]
[tex]d=\sqrt{(-3)^{2}+(0)^{2}}[/tex]
[tex]dBC2=3\ units[/tex]  Step 6Find the distance AC3[tex]A(1,-1)\ C3(0,4)[/tex]Substitute the values in the formula[tex]d=\sqrt{(4+1)^{2}+(0-1)^{2}}[/tex]
[tex]d=\sqrt{(5)^{2}+(-1)^{2}}[/tex]
[tex]dAC3=\sqrt{26}\ units[/tex]
Step 7Find the distance BC3[tex]B(3,2)\ C3(0,4)[/tex]Substitute the values in the formula[tex]d=\sqrt{(4-2)^{2}+(0-3)^{2}}[/tex]
[tex]d=\sqrt{(2)^{2}+(-3)^{2}}[/tex]
[tex]dBC3=\sqrt{13}\ units[/tex]
we know thatIf the length sides of the triangle satisfy the Pythagoras Theorem. then the triangle is a right triangleThe formula of the Pythagoras Theorem is equal to[tex]c^{2} =a^{2}+b^{2}[/tex]where c is the hypotenuse (the greater side)a and b are the legs of the triangle Step 8Verify if the triangle ABC1 is a right trianglewe have[tex]dAB=\sqrt{13}\ units[/tex]
[tex]dAC1=\sqrt{10}\ units[/tex]
[tex]dBC1=3\ units[/tex]
Applying Pythagoras theorem[tex]AB^{2}=AC1^{2}+BC1^{2}[/tex][tex]\sqrt{13}^{2} =\sqrt{10}^{2}+3^{2}[/tex][tex]13 =10+9[/tex][tex]13 =19[/tex] --------> is not truethereforethe triangle ABC1 is not a right triangleStep 9Verify if the triangle ABC2 is a right trianglewe have[tex]dAB=\sqrt{13}\ units[/tex]
[tex]dAC2=2\ units[/tex]
[tex]dBC2=3\ units[/tex]
Applying Pythagoras theorem[tex]AB^{2}=AC2^{2}+BC2^{2}[/tex][tex]\sqrt{13}^{2} =2^{2}+3^{2}[/tex][tex]13 =4+9[/tex][tex]13 =13[/tex] --------> is truethereforethe triangle ABC2 is a right triangleStep 10Verify if the triangle ABC3 is a right trianglewe have[tex]dAB=\sqrt{13}\ units[/tex]
[tex]dAC3=\sqrt{26}\ units[/tex]
[tex]dBC3=\sqrt{13}\ units[/tex]
Applying Pythagoras theorem[tex]AC3^{2}=AB^{2}+BC3^{2}[/tex][tex]\sqrt{26}^{2} =\sqrt{13}^{2}+\sqrt{13}^{2}[/tex][tex]26 =13+13[/tex][tex]26=26[/tex] --------> is truethereforethe triangle ABC3 is a right trianglethereforethe answer is[tex]C(0,2)\ Not\ a\ right\ triangle\\C(3,-1)\ A\ right\ triangle\\C(0,4)\ A\ right\ triangle[/tex]