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Q.

The combined equation of two sides of an equilateral tringle is $x^{2} - 3 y^{2} - 2 x + 1 = 0 .$ If the length of a side of the triangle is 4 then the equation of the third side is:

(this question has multiple options)

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a

x = 2 \sqrt{3} + 1

b

y = 2 \sqrt{3} + 1

c

x + 2 \sqrt{3} = 1

d

x = 2 \sqrt{3}

answer is A.

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Detailed Solution

Given the equation is,

x^{2} - 3 y^{2} - 2 x + 1 = 0

Here,

x^{2} - 2 x + 1 = 3 y^{2}

(x - 1)^{2} = 3 y^{2}

[since

(a - b)^{2} = a^{2} - 2 ab + b^{2}

]

x - 1

= \pm \sqrt{3} y

Therefore, the sides' equation can be given as,

x - \sqrt{3} y = 1

and

x + \sqrt{3} y = 1

They come together at (1, 0). Because of this, one of the vertices = (1, 0).
Here, equation for the third side will be parallel to the y-axis as well as perpendicular to the x-axis.
Here,

x - \sqrt{3} y = 1

and

x + \sqrt{3} y = 1

both have the same x-intercept while having equal and opposite y intercepts.
Both are equally inclined to the positive x axis; one is in the clockwise direction and the other is the anticlockwise direction.
Also,

x - \sqrt{3} y = 1

as the line's representation

x + \sqrt{3} y = 1

with relation to the x axis.
Thus, the third line will have the following format:

x = c .

The vertex's (1, 0) distance from the line above will now be

\frac{| 1 - c |}{1} = asin 60^{\circ}

\frac{| 1 - c |}{1} = 4 \sin 60^{\circ}

(a=4, side of triangle)

\frac{| 1 - c |}{1} = 2 \sqrt{3}

|1 - c| = 2 \sqrt{3}

c - 1 = 2 \sqrt{3}

c = 2 \sqrt{3} + 1

and

c = - 2 \sqrt{3} + 1

Thus, the equations of the third side are

x = 2 \sqrt{3} + 1

and

x + 2 \sqrt{3} = 1

.
Hence, the correct options are 1 and 3.

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