Question

If $(4,3)$ and $(-4,3)$ are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

• A. $(0,-3-4\sqrt{3})$
• B. $(0,3+4\sqrt{3})$
• C. $(0,3-4\sqrt{3})$
• D. $(0,-3+4\sqrt{3})$

Answer 1

The correct option is C $(0,3-4\sqrt{3})$

Let the given equilateral triangle be $ABC$ with vertices

$A(x_1,y_1)=(4,3),B(x_2,y_2)=(-4,3)\&C(x_3,y_3)=(x,y)$ .

Since the ordinates of $A$ & $B$ are same,

$\therefore AB\parallel X-$axis.

So $AB=$ Difference of the abscissae $=4-(-4)$ units = $8$ units.

i.e $AB=BC=AC=8$ units.

The abscissae of $A$ & $B$ are equal but opposite in sign.

$\therefore A\&B$ are equidistant from the $Y-$axis.

$\Delta$ ABC is equilateral.

So, in this case, $C$ lies on the $Y-$axis.

$C$ may lie on the either side of $AB$.

But the origin lies in the interior of the $\Delta$.

So $C$ will lie downwards $AB$ & on the $Y-$axis. i.e the co-ordinates of $C$ is $(0,y).$

Applying the distance formula

$d=\sqrt{{(x_3-x_2)}^2+{(y_3-y_2)}^2}$

$BC=\sqrt{{(0+4)}^2+{(y-3)}^2}=\sqrt{y^2-6y+25}=8\Rightarrow y^2-6y-39=0$ and

$\Rightarrow y=(3+4\sqrt{3})$ which is positive & we reject it.

Or $y=(3-4\sqrt{3})$ which is negative & we accept it.

$\therefore$ The co-ordinates of $C$ are $(0,3-4\sqrt{3})$