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Troyanec [42]
4 years ago
10

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)
Mathematics
2 answers:
Naily [24]4 years ago
7 0

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

Scorpion4ik [409]4 years ago
6 0
If you graph 4 separate triangles with the four different points, (2,3) and (1,5) form a right triangle.
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