First slide

Equation of line in Straight lines

Question

ABCD is a square whose vertices A (0,0) B (2,0), C (2,2), D (0, 2). This square is rotated in the xy-plane with an angle of $30^{0}$ in anticlockwise direction about an axis passing through the vertex A, the equation of the diagonal BD of this rotated square is

Moderate

A

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

$(1 + \sqrt{3}) x - (1 - \sqrt{2}) y = 2$
C

$(2 - \sqrt{3}) x + y = 2 (\sqrt{3} - 1)$
D

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

Solution

New coordinates of
$B = (0 + 2 (\frac{\sqrt{3}}{2}), 0 + 2 (\frac{1}{2})) = (\sqrt{3}, 1)$
$C = (0 + 2 (- \frac{1}{2}), 0 + 2 (\frac{\sqrt{3}}{2})) = (- 1, \sqrt{3})$
New equation of BD is $y - \sqrt{3} = \frac{\sqrt{3} - 1}{- (\sqrt{3} + 1)} (x + 1)$
$y - \sqrt{3} = - (2 - \sqrt{3}) (x + 1)$
$= - (2 - \sqrt{3}) x - 2 + \sqrt{3}$
i.e., $(2 - \sqrt{3}) x + y = 2 (\sqrt{3} - 1)$

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