Question

In △ABC, the coordinates of vertices A and B are A(1,−1) and B(3,2).

Answer 1

C90,2) - Not a right triangle

C(3,-1) - is a right triangle ( its slope is a negative inverse )

C(0,4) - is a right triangle ( its slope is a negative inverse )

Answer 2

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

In this problem we have

$A(1,-1)\ B(3,2)\ C1(0,2)\ C2(3,-1)\ C3(0,4)$

Step 1

Find the distance AB

$A(1,-1)\ B(3,2)$

Substitute the values in the formula

$d=\sqrt{(2+1)^{2}+(3-1)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

Step 2

Find the distance AC1

$A(1,-1)\ C1(0,2)$

Substitute the values in the formula

$d=\sqrt{(2+1)^{2}+(0-1)^{2}}$

$d=\sqrt{(3)^{2}+(-1)^{2}}$

$dAC1=\sqrt{10}\ units$

Step 3

Find the distance BC1

$B(3,2)\ C1(0,2)$

Substitute the values in the formula

$d=\sqrt{(2-2)^{2}+(0-3)^{2}}$

$d=\sqrt{(0)^{2}+(-3)^{2}}$

$dBC1=3\ units$

Step 4

Find the distance AC2

$A(1,-1)\ C2(3,-1)$

Substitute the values in the formula

$d=\sqrt{(-1+1)^{2}+(3-1)^{2}}$

$d=\sqrt{(0)^{2}+(2)^{2}}$

$dAC2=2\ units$

Step 5

Find the distance BC2

$B(3,2)\ C2(3,-1)$

Substitute the values in the formula

$d=\sqrt{(-1-2)^{2}+(3-3)^{2}}$

$d=\sqrt{(-3)^{2}+(0)^{2}}$

$dBC2=3\ units$  

Step 6

Find the distance AC3

$A(1,-1)\ C3(0,4)$

Substitute the values in the formula

$d=\sqrt{(4+1)^{2}+(0-1)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC3=\sqrt{26}\ units$

Step 7

Find the distance BC3

$B(3,2)\ C3(0,4)$

Substitute the values in the formula

$d=\sqrt{(4-2)^{2}+(0-3)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC3=\sqrt{13}\ units$

we know that

If the length sides of the triangle satisfy the Pythagoras Theorem. then the triangle is a right triangle

The formula of the Pythagoras Theorem is equal to

$c^{2} =a^{2}+b^{2}$

where

c is the hypotenuse (the greater side)

a and b are the legs of the triangle

Step 8

Verify if the triangle ABC1 is a right triangle

we have

$dAB=\sqrt{13}\ units$

$dAC1=\sqrt{10}\ units$

$dBC1=3\ units$

Applying Pythagoras theorem

$AB^{2}=AC1^{2}+BC1^{2}$

$\sqrt{13}^{2} =\sqrt{10}^{2}+3^{2}$

$13 =10+9$

$13 =19$ --------> is not true

therefore

the triangle ABC1 is not a right triangle

Step 9

Verify if the triangle ABC2 is a right triangle

we have

$dAB=\sqrt{13}\ units$

$dAC2=2\ units$

$dBC2=3\ units$

Applying Pythagoras theorem

$AB^{2}=AC2^{2}+BC2^{2}$

$\sqrt{13}^{2} =2^{2}+3^{2}$

$13 =4+9$

$13 =13$ --------> is true

therefore

the triangle ABC2 is a right triangle

Step 10

Verify if the triangle ABC3 is a right triangle

we have

$dAB=\sqrt{13}\ units$

$dAC3=\sqrt{26}\ units$

$dBC3=\sqrt{13}\ units$

Applying Pythagoras theorem

$AC3^{2}=AB^{2}+BC3^{2}$

$\sqrt{26}^{2} =\sqrt{13}^{2}+\sqrt{13}^{2}$

$26 =13+13$

$26=26$ --------> is true

therefore

the triangle ABC3 is a right triangle

therefore

the answer is

$C(0,2)\ Not\ a\ right\ triangle\\C(3,-1)\ A\ right\ triangle\\C(0,4)\ A\ right\ triangle$