Question

One of the vertices of an equilateral triangle is $(2,3)$ and the equation of its opposite side is $x+y-2=0$. The area of triangle is?

• A. $3\sqrt{3}$
• B. $6\sqrt{3}$
• C. $\frac{3\sqrt{3}}{2}$
• D. $\sqrt{6}$

Answer 1

The correct option is C $\frac{3\sqrt{3}}{2}$

In equiletral triangle,

Length of altitude$=a\sin {60}^{\circ }$

Let altitude$A$

So, $A=\frac{a\sqrt{3}}{2}$

$⟹a=\frac{2A}{\sqrt{3}}$

Now a$=$ side

And Area$=(\frac{a\sqrt{3}}{2}{)}^2=\frac{A^2}{\sqrt{3}}$

And $A=$ perpendicular distance from$(2,3)$ to $x+y-2=0$

So, $A=|\frac{2+3-2}{\sqrt{1^2+1^2}}|=\frac{3}{\sqrt{2}}⟹$ Area$=\frac{A^2}{\sqrt{3}}=\frac{3\sqrt{3}}{2}$