Show that the triangle whose vertices are (8, 2), (5, – 3) and (0, 0

1.	Show that the triangle whose vertices are (8, 2), (5, – 3) and (0, 0) is an isosceles.
Answer» Let, the given vertices be A(x₁, y₁) = (8, 2), B(x₂, y₂) = (5, – 3) and C(x₃, y₃) = (0, 0) ∴ AB = \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) = \(\sqrt{(8-5)^2+(2+3)^2}\) =\(\sqrt{34}\) units. C =\(\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}\) = \(\sqrt{(5-0)^2+(-3-0)^2}\) = \(\sqrt{34}\) units. ∴ AB = BC (⇒ ∠ACB = ∠BAC). Hence, ∆ABC is isosceles.

Show that the triangle whose vertices are (8, 2), (5, – 3) and (0, 0) is an isosceles.

Answer»

Let, the given vertices be A(x₁, y₁) = (8, 2),

B(x₂, y₂) = (5, – 3) and C(x₃, y₃) = (0, 0)

∴ AB = \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

= \(\sqrt{(8-5)^2+(2+3)^2}\) =\(\sqrt{34}\) units.

C =\(\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}\)

= \(\sqrt{(5-0)^2+(-3-0)^2}\) = \(\sqrt{34}\) units.

∴ AB = BC (⇒ ∠ACB = ∠BAC).

Hence, ∆ABC is isosceles.