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.
C(0,2)
C(3,−1)
C(0,4)

Answer 1

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=√(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=√(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=√(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=√(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=√(13)\ units`

`dAC1=√(10)\ units`

`dBC1=3\ units`

Applying Pythagoras theorem

`AB^(2)=AC1^(2)+BC1^(2)`

`√(13)^(2) =√(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=√(13)\ units`

`dAC2=2\ units`

`dBC2=3\ units`

Applying Pythagoras theorem

`AB^(2)=AC2^(2)+BC2^(2)`

`√(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=√(13)\ units`

`dAC3=√(26)\ units`

`dBC3=√(13)\ units`

Applying Pythagoras theorem

`AC3^(2)=AB^(2)+BC3^(2)`

`√(26)^(2) =√(13)^(2)+√(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`