Answer:
Correct options are 2, 5 and 7.
Step-by-step explanation:
Consider the given vertices of triangle are A(-3,-3), B(-3,2) and C(1,2).
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, we get
[tex]AB=\sqrt{(-3-(-3))^2+(2-(-3))^2}[/tex]
[tex]AB=\sqrt{(0)^2+(5)^2}[/tex]
[tex]AB=\sqrt{25}[/tex]
[tex]AB=5[/tex]
Similarly,
[tex]BC=\sqrt{(1-(-3))^2+(2-2)^2}=4[/tex]
[tex]AC=\sqrt{(1-(-3))^2+(2-(-3))^2}=\sqrt{16+25}=\sqrt{41}[/tex]
From the above calculation it is clear that AC>AB and AC>BC.
According to Pythagoras theorem, in a right angle triangle, the square of largest side is equal to the sum of squares of two small sides.
[tex]hypotenuse^2=base^2+perpendicular^2[/tex]
[tex]AC^2=(\sqrt{41})^2=41[/tex]
[tex]AB^2+BC^2=(5)^2+4^2=24+16=41=AC^2[/tex]
So, given triangle is a right angle triangle and AC is its hypotenuse.
Therefore, the correct options are 2, 5 and 7.
