Let a square with side length - p - and making an angle of - with x - axis- has one vertex at origin- If 0-2- then the equation of the diagonals of the square isA- x1-tan-y1-tan-py1-tan-x1-tan-B- xcos-sin-ycos-sin-py cos-sin- x cos-sin-C- xcos-sin-ycos-sin-py cos-sin- x cos-sin-D- x1-tan-4-y1-tan-4-p-ytan-4-xtan-4-

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Condition for Two Lines to Be Perpendicular

Let a square ...

Question

Let a square with side length $^{'} p^{'}$ and making an angle of $θ$ with $x -$ axis, has one vertex at origin. If $0 < θ < \frac{π}{2}$ , then the equation of the diagonals of the square is

x (1 + tan θ) - y (1 + tan θ) = p y (1 + tan θ) = x (1 - tan θ)

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x (cos θ + sin θ) - y (cos θ + sin θ) = p y (cos θ + sin θ) = x (cos θ + sin θ)

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x (cos θ - sin θ) + y (c o s θ + sin θ) = p y (cos θ - sin θ) = x (cos θ + sin θ)

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x (1 + tan (\frac{π}{4} + θ) - y (1 + tan (\frac{π}{4} + θ) = p y (tan (\frac{π}{4} + θ)) = x (tan (\frac{π}{4} - θ))

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Solution

The correct option is C $x (cos θ - sin θ) + y (c o s θ + sin θ) = p y (cos θ - sin θ) = x (cos θ + sin θ)$
Equation of the diagonal with vertex at origin -
$y - 0 = m (x - 0)$
$m = tan (\frac{π}{4} + θ)$
$\Rightarrow y (1 - tan θ) = x (1 + tan θ) \Rightarrow y (cos θ - sin θ) = x (cos θ + sin θ)$
Since, diagonals of a square are perpendicular to each other therefore,
$m = \frac{- 1}{tan (\frac{π}{4} + θ)}$
$(y - p sin θ) = \frac{- 1}{tan (\frac{π}{4} + θ)} (x - p cos θ) x (cos θ - sin θ) + y (cos θ + sin θ) = p$
Alternet:
perpendicular distance is $\frac{p}{\sqrt{2}}$ and angle makes with $x -$ axis is $(θ + 45 °)$
so the equation will be $x cos (θ + 45 °) + y sin (θ + 45 °) = \frac{p}{\sqrt{2}} \Rightarrow x (cos θ - sin θ) + y (cos θ + sin θ) = p$

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