Question

Two vertices of a right triangle are located at (1, 3) and (2, 5) .Select each ordered pair that could be the coordinates of the third vertex. A(1,5) B(2,3) C(3,3) D(3,5)

Answer 1

a and b

Explanation:

i took the quiz :)

Answer 2

Option A and B are correct.

Explanation:

Two vertices of a right triangle are located at A(1, 3) and B(2, 5)

We have to choose the third vertex from the options. We have to check each option.

Using Distance formula:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Pythagorous Identity:

`H^2=P^2+B^2`

We will find the length of each side and then apply pyhtagoreous property.

If Pythagorean theorem follow then it could be third vertex else not

Option A: C(1,5)

`AB=√((2-1)^2+(5-3)^2)=√(5)`

`AC=√((1-1)^2+(5-3)^2)=√(4)`

`BC=√((2-1)^2+(5-5)^2)=√(1)`

`AB^2=AC^2+BC^2`

`(√(5))^2=(√(4))^2+(√(1))^2`

`5=4+1`

`5=5`

TRUE

Option B: C(2,3)

`AB=√((2-1)^2+(5-3)^2)=√(5)`

`AC=√((1-2)^2+(3-3)^2)=√(1)`

`BC=√((2-2)^2+(5-3)^2)=√(4)`

`AB^2=AC^2+BC^2`

`(√(5))^2=(√(4))^2+(√(1))^2`

`5=4+1`

`5=5`

TRUE

Option C: C(3,3)

`AB=√((2-1)^2+(5-3)^2)=√(5)`

`AC=√((1-3)^2+(3-3)^2)=√(4)`

`BC=√((2-3)^2+(5-3)^2)=√(5)`

`AB^2=AC^2+BC^2`

`(√(5))^2=(√(4))^2+(√(5))^2`

`5=4+5`

`5\\eq 9`

FALSE

Option D: C(3,5)

`AB=√((2-1)^2+(5-3)^2)=√(5)`

`AC=√((1-3)^2+(3-5)^2)=√(8)`

`BC=√((2-3)^2+(5-5)^2)=√(1)`

`AB^2=AC^2+BC^2`

`(√(5))^2=(√(8))^2+(√(1))^2`

`5=8+1`

`5\\eq 9`

FALSE