Question

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

• A. $x=2\sqrt{3}+1$
• B. $y=2\sqrt{3}+1$
• C. $x+2\sqrt{3}=1$
• D. $x=2\sqrt{3}$

Answer 1

Answer: (A), (C)

The correct options are
A $x+2\sqrt{3}=1$
C $x=2\sqrt{3}+1$

$x^2-3y^2-2x+1=0$

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

$x-1=\pm \sqrt{3}y$

Hence the equation of the sides are

$x-\sqrt{3}y=1$ and $x+\sqrt{3}y=1$

They intersect at $(1,0)$. Hence one of the vertex will be $(1,0)$.

Now we can clearly observe that the equation of the third side will be perpendicular to x-axis and parallel to y-axis since

$x-\sqrt{3}y=1$ and $x+\sqrt{3}y=1$ are equally inclined to positive x axis- one in clockwise sense and another in anticlockwise sense, and both have the same x-intercept while equal and opposite y intercept. In other words we can imagine $x-\sqrt{3}y=1$ as the image of the line $x+\sqrt{3}y=1$ with respect to x axis.

Hence the third line will be of the form $x=c$.

Now distance of the vertex $(1,0)$ from the above line will be

$=asin{60}^0$

$=4sin{60}^0$

$=2\sqrt{3}$.

$=\frac{|1-c|}{1}$

Or

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

Hence

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

$c=2\sqrt{3}+1$ and $c=-2\sqrt{3}+1$

Hence the corresponding equations are

$x=2\sqrt{3}+1$ and $x+2\sqrt{3}=1$.