Question

One diagonal of a square is the portion of the line $\sqrt{3}x+y=2\sqrt{3}$ intercepted by the axes. Then extremity of the diagonals are

• A. $(1+\sqrt{3},\sqrt{3}-1)$
• B. $(1+\sqrt{3},\sqrt{3}+1)$
• C. $(1-\sqrt{3},\sqrt{3}-1)$
• D. $(1-\sqrt{3},\sqrt{3}+1)$

Answer 1

Answer: (B), (C)

The correct options are
B $(1+\sqrt{3},\sqrt{3}+1)$
C $(1-\sqrt{3},\sqrt{3}-1)$

Given line

$\sqrt{3}x+y=2\sqrt{3}$

$\frac{x}{2}+\frac{y}{2\sqrt{3}}=1$

On comparing above eq with intercept form of line

$A(2,0)$ and $B(0,2\sqrt{3})$

SO mid point of one diagonal is $(1,\sqrt{3})$

Length of diagonal $=\sqrt{(2-0{)}^2+(0-2\sqrt{3}{)}^2}=\sqrt{4+12}=4$

Slope of diagonal $=-\frac{2\sqrt{3}}{2}=-\sqrt{3}$

Slope of line perpendicular to diagonal $\frac{1}{\sqrt{3}}$

Eq of other diagonal which is perpendicular to the diagonal and passing through point is

$\frac{x-1}{\frac{\sqrt{3}}{4}}=\frac{y-\sqrt{3}}{\frac{1}{4}}=r$

Substitute $r=\pm 4$

$\frac{x-1}{\frac{\sqrt{3}}{4}}=\pm 4$

$x-1=\pm 4\times \frac{\sqrt{3}}{4}$

$x=\pm \sqrt{3}+1$

$x=(1+\sqrt{3}),(1-\sqrt{3})$

$y=(\sqrt{3}+1),(\sqrt{3}-1)$

coordinates of extremity $(1+\sqrt{3},\sqrt{3}+1)$ and $1+\sqrt{3},\sqrt{3}-1$