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

Answer:

Option A and B are correct.

Step-by-step 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:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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)

$AB=\sqrt{(2-1)^2+(5-3)^2}=\sqrt{5}$

$AC=\sqrt{(1-1)^2+(5-3)^2}=\sqrt{4}$

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

$AB^2=AC^2+BC^2$

$(\sqrt{5})^2=(\sqrt{4})^2+(\sqrt{1})^2$

$5=4+1$

$5=5$

TRUE

Option B: C(2,3)

$AB=\sqrt{(2-1)^2+(5-3)^2}=\sqrt{5}$

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

$BC=\sqrt{(2-2)^2+(5-3)^2}=\sqrt{4}$

$AB^2=AC^2+BC^2$

$(\sqrt{5})^2=(\sqrt{4})^2+(\sqrt{1})^2$

$5=4+1$

$5=5$

TRUE

Option C: C(3,3)

$AB=\sqrt{(2-1)^2+(5-3)^2}=\sqrt{5}$

$AC=\sqrt{(1-3)^2+(3-3)^2}=\sqrt{4}$

$BC=\sqrt{(2-3)^2+(5-3)^2}=\sqrt{5}$

$AB^2=AC^2+BC^2$

$(\sqrt{5})^2=(\sqrt{4})^2+(\sqrt{5})^2$

$5=4+5$

$5\neq 9$

FALSE

Option D: C(3,5)

$AB=\sqrt{(2-1)^2+(5-3)^2}=\sqrt{5}$

$AC=\sqrt{(1-3)^2+(3-5)^2}=\sqrt{8}$

$BC=\sqrt{(2-3)^2+(5-5)^2}=\sqrt{1}$

$AB^2=AC^2+BC^2$

$(\sqrt{5})^2=(\sqrt{8})^2+(\sqrt{1})^2$

$5=8+1$

$5\neq 9$

FALSE

Answer 2

If you graph 4 separate triangles with the four different points, (2,3) and (1,5) form a right triangle.