Question 19

If two vertices of an equilateral triangle be (0,0) and (3, $\sqrt{3}$), find the third vertex using distance formula.

Given,

A = (0, 0)

B =(3, $\sqrt{3}$)

Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Length of AB:

$= \sqrt{(3 - 0)^2 + (\sqrt{3} - 0)^2} \\[1em] = \sqrt{(3)^2 + (\sqrt{3})^2} \\[1em] = \sqrt{9 + 3} \\[1em] = \sqrt{12}$

Let the third vertex be C(x, y)

Since the triangle is equilateral, the distance from C to A and C to B must also be $\sqrt{12}$.

Length of AC = Length of AB

$\Rightarrow \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{12} \\[1em] \text{Squaring on both sides} \\[1em] \Rightarrow x^2 + y^2 = 12 ……..(1)$

Length of BC = Length of AB

$\Rightarrow \sqrt{(x - 3)^2 + (y - \sqrt{3})^2} = \sqrt{12} \\[1em] \text{Squaring on both sides} \\[1em] \Rightarrow (x - 3)^2 + (y - \sqrt{3})^2 = 12 \\[1em] \Rightarrow x^2 - 6x + 9 + y^2 - 2\sqrt{3}y + 3 = 12 \\[1em] \Rightarrow x^2 + y^2 - 6x - 2\sqrt{3}y + 12 = 12 \\[1em] \Rightarrow x^2 + y^2 - 6x - 2\sqrt{3}y = 0 ……….(2) \\[1em] \text{Substitute equation (1) in (2)} \\[1em] \Rightarrow 12 - 6x - 2\sqrt{3}y = 0 \\[1em] \text{Divide by 2} \\[1em] \Rightarrow 6 - 3x - \sqrt{3}y = 0 \\[1em] \Rightarrow \sqrt{3}y = 6 - 3x \\[1em] \Rightarrow y = \dfrac{6 - 3x}{\sqrt{3}} = 2\sqrt{3} - x\sqrt{3} \\[1em] \Rightarrow y = \sqrt{3}(2 - x)$

Substitute the value of y in equation (1),

x² + ($\sqrt{3}$(2 - x))² = 12

x² + 3(4 - 4x + x²) = 12

x² + 12 - 12x + 3x² = 12

4x² - 12x = 0

4x(x - 3) = 0

∴ x = 0 or x = 3

If x = 0, then

y = $\sqrt{3}(2 - 0) = 2\sqrt{3}$

Vertex C = (0, $2\sqrt{3}$)

If x = 3, then

y = $\sqrt{3}(2 - 3) = -\sqrt{3}$

Vertex C = (3, $-\sqrt{3}$).

Hence, third vertex = (0, $2\sqrt{3}$) and (3, $-\sqrt{3}$).