Show that the points A(-5,6), B(3, 0) and C( 9, 8) are

Q) Show that the points A(-5,6), B(3, 0) and C( 9, 8) are the vertices of an isosceles triangle.

Ans:

For a triangle to be an isosceles triangle, its any 2 sides should be equal.

Let’s start by calculating length of its sides:

We know that the distance between two points P (X₁, Y₁) and Q (X₂, Y₂) is given by:

PQ = $\sqrt {(\times_2 - \times _1)^2 + (Y_2 - Y_1)^2}$

We have coordinates as A(-5,6), B(3, 0) and C( 9, 8)

∴ AB = $\sqrt{(3-(- 5))^2 + (0 - 6)^2}$

∴ AB = $\sqrt{64 +36}$

∴ AB = $\sqrt{100}$

∴ AB = 10

Similarly, BC = $\sqrt{(9 - 3)^2 + (8 - 0)^2}$

∴ BC = $\sqrt{36 + 64}$

∴ BC = $\sqrt{100}$

∴ BC = 10

Here, since AB = BC

Therefore, A, B and C are vertices of an isosceles triangle.

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