$\begin {array} {1 1} (A)\;\large\frac{2}{\sqrt 3} & \quad (B)\;\large\frac{2}{ 3} \\ (C)\;\large\frac{\sqrt 2}{ 3} & \quad (D)\;\sqrt{\large\frac{2}{3}} \end {array}$

- Length of the perpendicular from $(x_1, y_1)$ to the line $ax+by+c=0$ is $ p = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$

Step 1 :

Let $a$ be the length of each side of the equilateral triangle.

The length of the perpendicular p from (2,-1) on $x+y-2=0$

$ \Rightarrow p = \bigg| \large\frac{2-1-2}{\sqrt{1^2+1^2}}\bigg|$

$ = \large\frac{1}{\sqrt 2}$

Step 2 :

Consider the $ \Delta $ ABD

$ \sin 60^{\circ} = \large\frac{p}{a}$

$ \Rightarrow \large\frac{\sqrt 3}{2} $$ = \large\frac{\Large\frac{1}{\sqrt 2}}{a}$

$ \Rightarrow a = \large\frac{2}{\sqrt 3}$$ \times \large\frac{1}{\sqrt 2}$

$ a = \large\frac{\sqrt 2}{3}$

hence the length of the side of the equilateral triangle is $ \large\frac{\sqrt 2}{3}$

Option D is the correct answer root 2 by root 3. Thanks

Ask Question

Tag:MathPhyChemBioOther

Take Test

...