If two vertices of an equlateral triangle are (0,0) and (3,√ 3) Find the third vertex Send - Maths - Coordinate Geometry

Question

If two vertices of an equlateral triangle are (0,0) and
(3,√ 3) Find the third vertex Send solution by meritnation
expert only Thanking you

Lalit Mehra · Accepted Answer

Suppose A(0, 0), B $(3, \sqrt{3})$ and C (x, y) be the vertices of the equilateral triangle ABC.

∴ AB = BC = CA (Sides of equilateral triangle are equal)

⇒ AB² = BC² = CA²

$\Rightarrow (3 - 0)^{2} + (\sqrt{3} - 0)^{2} = (x - 3)^{2} + (y - \sqrt{3})^{2} = (x - 0)^{2} + {(y - 0)}^{2} \Rightarrow 9 + 3 = {(x - 3)}^{2} + (y - \sqrt{3})^{2} = x^{2} + y^{2} \Rightarrow 12 = x^{2} - 6 x + 9 + y 2 - 2 \sqrt{3} y + 3 = x^{2} + y^{2} ∴ x^{2} + y^{2} = 12 ... (1) and x^{2} + y^{2} - 6 x - 2 \sqrt{3} y + 12 = x^{2} + y^{2} ∴ - 6 x - 2 \sqrt{3} y + 12 = 0 \Rightarrow y = \frac{12 - 6 x}{2 \sqrt{3}} \Rightarrow y = 2 \sqrt{3} - \sqrt{3} x ... (2)$

From (1) and (2), we have

$x^{2} + (2 \sqrt{3} - \sqrt{3} x)^{2} = 12$

∴ x² + 12 + 3x² – 12x = 12

⇒ 4x² – 12x = 0

⇒ 4x (x – 3) = 0

⇒ x = 0 or x = 3.

When x = 0,

$y = 2 \sqrt{3} - \sqrt{3} \times 0 = 2 \sqrt{3} - 0 = 2 \sqrt{3}$

When x = 3,

$y = 2 \sqrt{3} - \sqrt{3} \times 3 = 2 \sqrt{3} - 3 \sqrt{3} = - \sqrt{3}$

Thus, the third vertex of the equilateral triangle is $(0, 2 \sqrt{3})$ or $(3, - \sqrt{3})$

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If two vertices of an equlateral triangle are (0,0) and (3,√ 3) . Find the third vertex. Send solution by meritnation expert only. Thanking you.