Question

$P$ is a point on the line $y+2x=1$ and $Q$ and $R$ are two points on the line $3y+6x=6$ such that $△PQR$ is an equilateral triangle. Then the area of triangle is

• A. $\frac{1}{5\sqrt{5}}$
• B. $\frac{1}{5\sqrt{3}}$
• C. $\frac{2}{3\sqrt{5}}$
• D. $\frac{3}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{1}{5\sqrt{3}}$

Given, $\Delta PQR$ is equilateral triangle.

Given Lines: $y+2x=1$ ; $3y+6x=6\Rightarrow y+2x=2$

As both lines have same slope thus given lines are parallel.

To find: Area of $\Delta PQR.$

For finding Area, we would do so by finding altitude of $\angle PQR$ which would be simply the perpendicular distance between both given lines because both lines are parallel.

So,

Distance formula for $\perp$ distance between two lines= $d=\frac{|c_1–c_2|}{\sqrt{a^2+b^2}}$

So distance between given lines:$\frac{|2-1|}{\sqrt{2^2+1^2}}\Rightarrow \frac{1}{\sqrt{5}}$=Altitude of $\Delta PQR$

Now, Area of equilateral triangle in terms of altitude ${}^{\prime }{h}^{\prime }=\frac{h^2}{\sqrt{3}}$

so Area of $\Delta PQR$=$\frac{(\frac{1}{\sqrt{5}}{)}^2}{\sqrt{3}}$$\Rightarrow \frac{1}{5\sqrt{3}}$

Hence,Option (B) is correct.