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 is

Equation of the diagonal with vertex at origin y 0=m(x 0) m =tan(π/4+θ) ⇒ y(1 tanθ)=x(1+tanθ) ⇒ y(cosθ sinθ)=x(cosθ+sinθ) Since, diagonals of a square are per ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

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)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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))

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $x cos (θ + 45 °) + y sin (θ + 45 °) = \frac{p}{\sqrt{2}} \Rightarrow x (cos θ - sin θ) + y (cos θ + sin θ) = p$

Suggest Corrections

Similar questions

Q. 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

Q. A square of side

a

lies above x - axis and has one vertex at the origin. The side passing through the origin makes an angle a

(0 < a < \frac{π}{4})

with positive direction of x - axis. The equation of its diagonal not passing through the origin is

Q. A square with each side equal to a lies above the

x

-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle

α (0 < α < π / 4)

with the positive direction of the

x

-axis. Equation of a diagonal of the square is

Q. A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle

α [0 < α < \frac{π}{4}]

with the positive direction of x-axis. The equation of its diagonal not passing through the origin is

Q. A square of side

a

lies above the x-axis and has one vertex at the origin. The side passing through it makes an angle

α (0 < α < \frac{π}{4})

with the positive direction of x-axis. The equation of its diagonal not through the origin

Join BYJU'S Learning Program

Let a square with side length ′p′ and making an angle of θ with x− axis, has one vertex at origin. If 0<θ<π2, then the equation of the diagonals of the square is

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