Question

The area of an equilateral triangle with the equation of base as x+y-2=0 and the opposite vertex with the coordinates (2, -1) is sq. units.

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{1}{3\sqrt{3}}$

Answer 1

The correct option is B $\frac{1}{2\sqrt{3}}$

Let 'p' be the length of the altitude i.e. the perpendicular distance from the point (2, -1) on the line x+y-2=0.
$\Rightarrow p=\left(\frac{2-1-2}{\sqrt{1^2+1^2}}\right|=\frac{1}{\sqrt{2}}\mathrm{Also},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{altitude}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{with}\mathrm{side}a\mathrm{is}\frac{\sqrt{3}a}{2}(\because \mathrm{Area}=\frac{1}{2}\times p\times a=\frac{\sqrt{3}{a}^2}{4})\Rightarrow \frac{\sqrt{3}a}{2}=\frac{1}{\sqrt{2}}\Rightarrow a=\sqrt{\frac{2}{3}}\Rightarrow \mathrm{Area}=\frac{\sqrt{3}{a}^2}{4}=\frac{\sqrt{3}}{4}\times \frac{2}{3}=\frac{1}{2\sqrt{3}}\mathrm{sq}.\mathrm{units}$