Question

The co-ordinates of vertices $P$ and $Q$ of an equilateral $\Delta$ are $(1,\sqrt{3})$ and $(0,0)$. Which of the following could be co-ordinates of $R$?

• A. $(1,2)$
• B. $(2,0)$
• C. $(1,\frac{\sqrt{3}}{2})$
• D. $(\sqrt{3},1)$

Answer 1

Answer: (B)

$(2,0)$

Given: $P(1,\sqrt{3}),Q(0,0)$ and $R(x,y)$ are the vertices of an equilateral $\Delta PQR$

To find: $R(x,y)=?$

We have $PQ=QR=RP$

Since $\Delta PQR$ is equilateral, we shall use the distance formula

$d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

So $PQ=\sqrt{{(1-0)}^2+{(\sqrt{3}-0)}^2}=\sqrt{4}$

So the triangle has sides $=2$

$QR=\sqrt{{(x-0)}^2+{(y-0)}^2}=\sqrt{x^2+y^2}$ and

$RP=\sqrt{{(x-1)}^2+{(y-\sqrt{3})}^2}=\sqrt{x^2{+y}^2-2x-2\sqrt{3}y+4}$

$\therefore \sqrt{x^2{+y}^2}=\sqrt{4}$

$\Rightarrow x^2{+y}^2=4$ ....... $\left(i\right)$ and

$\sqrt{x^2{+y}^2-2x-2\sqrt{3}y+4}=\sqrt{x^2{+y}^2}$

$\Rightarrow x^2{+y}^2-2x-2\sqrt{3}y+4=x^2{+y}^2$

$\Rightarrow x=2-\sqrt{3}y$ ...... $\left(ii\right)$

Substituting $x=2-\sqrt{3}y$ in $\left(i\right),$ we get

${(2-\sqrt{3}y)}^2+y^2=4$

$\Rightarrow y(y-\sqrt{3})=0$

$\Rightarrow y=(0,\sqrt{3}).$ $Substitutingy=0in\left(i\right),$

$x^2=4\Rightarrow x=\pm 2$

So $R(x,y)=(2,0),(-2,0)$

If $R(x,y)=(-2,0),$ then

$RP=\sqrt{{(-2-1)}^2+{(0+\sqrt{3})}^2}$

$=\sqrt{12}\ne 2$

So we reject this value.

Again substituting $y=\sqrt{3}$ in $\left(i\right),$ we get

$x^2+3=0\Rightarrow x=\sqrt{-3}\notin R$

So we reject this value.

$\therefore R(x,y)=(2,0)$