Question

In the $\Delta ABC$, the co-ordinate of $B$ are $(0,0),AB=2,\angle ABC=\frac{\pi }{3}$ and middle point of $BC$ has the co-ordinates $(2,0)$. The coordinates of point A of the triangle is

• A. $(\frac{1}{2},\frac{\sqrt{3}}{2})$
• B. $(\frac{5}{3},\frac{1}{\sqrt{3}})$
• C. $(\frac{4+\sqrt{3}}{3},\frac{1}{2})$
• D. $(1,\sqrt{3})$

Answer 1

Answer: (D)

$(1,\sqrt{3})$

REF.Image.

middle point of BC lies an

X and B is an origin

$\Rightarrow$ C lies an x axis

D is mid point of BC.

$\Rightarrow (\frac{x+0}{2},\frac{0+0}{2})=(2,0)$

$\Rightarrow x=4$

$\therefore$ BC = 4 units

Given AB = 2 units

equation of $AB\Rightarrow y=mx$

$m=tan\theta =tan\left(\pi /3\right)=\sqrt{3}$

$\Rightarrow y=\sqrt{3x}\to$ equation of AB

$AB=\sqrt{(\alpha -0{)}^2+(\beta -0{)}^2}=\sqrt{{\alpha }^2+{\beta }^2}=2$ (Given)

$\Rightarrow {\alpha }^2+{\beta }^2=4...\left(1\right)$

$(\alpha ,\beta )$ lies on AB

$\Rightarrow \beta =\sqrt{3}\alpha$

$\Rightarrow {\beta }^2=3{\alpha }^2...\left(2\right)$

From (1) & (2)

${\alpha }^2+3{\alpha }^2=4\Rightarrow {\alpha }^2=1\Rightarrow \alpha =\pm 1$

since $\theta$= positive , $\alpha =+1$

$\therefore \beta =\sqrt{3}\alpha =\sqrt{3}\times \left(1\right)=\sqrt{3}$

$\therefore$ coordinates of $A=(1,\sqrt{3})$