Question

In figure, $△ABC$ is an equilateral triangle with coordinate of B and C as $B(1,0)$ and $C(5,0)$. Find the coordinate of vertex A.

• A. $A(3,2\sqrt{3})$
• B. $A(3,\sqrt{3})$
• C. $A(2,2\sqrt{3})$
• D. none of the above

Answer 1

The correct option is A $A(3,2\sqrt{3})$

The ordinates of both $B$ & $C$ are $0$.

So, $BC=$ (abscissa of B$-$abscissa of C)$=(5-1)=4units=a$

The side of equilateral $\Delta ABC.$

$\therefore$ Its height$=\frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{2}\times 4=2\sqrt{3}$ units.

Also, the mid point of $BC=(\frac{5+1}{2},\frac{0+0}{2})=(3,0).$

Now, $\Delta ABC$ is equilateral and $BC$ is on the x-axis.

So, the abscissa of A will be the mid point of BC.

The ordinate of A will be the height of $\Delta ABC$ from the mid point of BC.

$\therefore$ The coordinates of A are $(3,2\sqrt{3})$.