Given:
The coordinates of ∆ABC are A(-3, 2), B(5, 8) & C(11, 0).
To find:
The type of the given triangle.
Solution:
Distance formula:
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using the distance formula, we get
[tex]AB=\sqrt{(5-(-3))^2+(8-2)^2}[/tex]
[tex]AB=\sqrt{(8)^2+(6)^2}[/tex]
[tex]AB=\sqrt{64+36}[/tex]
[tex]AB=\sqrt{100}[/tex]
[tex]AB=10[/tex]
Similarly,
[tex]BC=\sqrt{(11-5)^2+(0-8)^2}[/tex]
[tex]BC=\sqrt{(6)^2+(8)^2}[/tex]
[tex]BC=\sqrt{36+64}[/tex]
[tex]BC=\sqrt{100}[/tex]
[tex]BC=10[/tex]
And,
[tex]AC=\sqrt{(11-(-3))^2+(0-2)^2}[/tex]
[tex]AC=\sqrt{(14)^2+(-2)^2}[/tex]
[tex]AC=\sqrt{196+4}[/tex]
[tex]AC=\sqrt{200}[/tex]
[tex]AC=10\sqrt{2}[/tex]
Two sides of the triangle are equal, i.e., [tex]AB=BC[/tex]. So, the triangle is an isosceles triangle.
Sum of square of two smaller side is
[tex]AB^2+BC^2=10^2+10^2[/tex]
[tex]AB^2+BC^2=100+100[/tex]
[tex]AB^2+BC^2=200[/tex]
[tex]AB^2+BC^2=AC^2[/tex]
Using the Pythagoras theorem, we can say that the given triangle is a right triangle.
Therefore, the correct options are B and F.