Question

find equation of the line

Answer 1

By normal form the equation of the line will be,

$\displaystyle\longrightarrow x\cos\omega+y\sin\omega=\sqrt2$

Case 1:- Let our line have negative slope, which implies the line makes an obtuse angle with positive x axis, i.e.,

$\longrightarrow\theta\in\left(\dfrac{\pi}{2},\ \pi\right)$

Relation between $\theta$ and $\omega$ is $\theta=\dfrac{\pi}{2}+\omega.$ So,

$\longrightarrow\dfrac{\pi}{2}+\omega\in\left(\dfrac{\pi}{2},\ \pi\right)$

$\longrightarrow\omega\in\left(0,\ \dfrac{\pi}{2}\right)$

Putting (x, y) = (√3, -1) in the equation of our line,

$\displaystyle\longrightarrow \sqrt3\cos\omega-\sin\omega=\sqrt2$

Dividing by 2,

$\displaystyle\longrightarrow\dfrac{\sqrt3}{2}\cos\omega-\dfrac{1}{2}\sin\omega=\dfrac{1}{\sqrt2}$

$\displaystyle\longrightarrow\sin\left(\dfrac{\pi}{3}\right)\cos\omega-\cos\left(\dfrac{\pi}{3}\right)\sin\omega=\dfrac{1}{\sqrt2}$

$\displaystyle\longrightarrow\sin\left(\dfrac{\pi}{3}-\omega\right)=\dfrac{1}{\sqrt2}$

But,

$\longrightarrow\omega\in\left(0,\ \dfrac{\pi}{2}\right)$

$\longrightarrow-\omega\in\left(-\dfrac{\pi}{2},\ 0\right)$

$\longrightarrow\dfrac{\pi}{3}-\omega\in\left(-\dfrac{\pi}{6},\ \dfrac{\pi}{3}\right)\ni\dfrac{\pi}{4}$

So,

$\displaystyle\longrightarrow\dfrac{\pi}{3}-\omega=\dfrac{\pi}{4}$

$\displaystyle\longrightarrow\omega=\dfrac{\pi}{12}$

Note that ω lies in 1st quadrant, so (cos ω) > 0, (sin ω) > 0.

Now,

$\longrightarrow\cos\omega=\sqrt{\dfrac{1+\cos(2\omega)}{2}}$

$\longrightarrow\cos\left(\dfrac{\pi}{12}\right)=\sqrt{\dfrac{1+\cos\left(\dfrac{\pi}{6}\right)}{2}}$

$\longrightarrow\cos\left(\dfrac{\pi}{12}\right)=\sqrt{\dfrac{1+\dfrac{\sqrt3}{2}}{2}}$

$\longrightarrow\cos\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2+\sqrt3}}{2}$

$\longrightarrow\cos\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{4+2\sqrt3}}{2\sqrt2}$

$\longrightarrow\cos\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt3+1}{2\sqrt2}$

and,

$\longrightarrow\sin\omega=\sqrt{\dfrac{1-\cos(2\omega)}{2}}$

$\longrightarrow\sin\left(\dfrac{\pi}{12}\right)=\sqrt{\dfrac{1-\cos\left(\dfrac{\pi}{6}\right)}{2}}$

$\longrightarrow\sin\left(\dfrac{\pi}{12}\right)=\sqrt{\dfrac{1-\dfrac{\sqrt3}{2}}{2}}$

$\longrightarrow\sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2-\sqrt3}}{2}$

$\longrightarrow\sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{4-2\sqrt3}}{2\sqrt2}$

$\longrightarrow\sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt3-1}{2\sqrt2}$

Hence the equation of our line becomes,

$\displaystyle\longrightarrow x\left(\dfrac{\sqrt3+1}{2\sqrt2}\right)+y\left(\dfrac{\sqrt3-1}{2\sqrt2}\right)=\sqrt2$

$\displaystyle\longrightarrow\underline{\underline{\left(\sqrt3+1\right)x+\left(\sqrt3-1\right)y=4}}$

Case 2:- Let our line have positive slope, which implies the line makes an acute angle with positive x axis, i.e.,

$\longrightarrow\theta\in\left(0,\ \dfrac{\pi}{2}\right)$

$\longrightarrow\dfrac{\pi}{2}+\omega\in\left(0,\ \dfrac{\pi}{2}\right)$

$\longrightarrow\omega\in\left(-\dfrac{\pi}{2},\ 0\right)$

Putting (x, y) = (√3, -1) in the equation of our line,

$\displaystyle\longrightarrow \sqrt3\cos\omega-\sin\omega=\sqrt2$

which gives, as we see earlier,

$\displaystyle\longrightarrow\sin\left(\dfrac{\pi}{3}-\omega\right)=\dfrac{1}{\sqrt2}$

But,

$\longrightarrow\omega\in\left(-\dfrac{\pi}{2},\ 0\right)$

$\longrightarrow-\omega\in\left(0,\ \dfrac{\pi}{2}\right)$

$\longrightarrow\dfrac{\pi}{3}-\omega\in\left(\dfrac{\pi}{3},\ \dfrac{5\pi}{6}\right)\ni\dfrac{3\pi}{4}$

So,

$\displaystyle\longrightarrow\dfrac{\pi}{3}-\omega=\dfrac{3\pi}{4}$

$\displaystyle\longrightarrow\omega=-\dfrac{5\pi}{12}$

Note that ω lies in 4th quadrant, so (cos ω) > 0, (sin ω) < 0.

Now,

$\longrightarrow\cos\left(-\dfrac{5\pi}{12}\right)=\sqrt{\dfrac{1+\cos\left(-\dfrac{5\pi}{6}\right)}{2}}$

$\longrightarrow\cos\left(-\dfrac{5\pi}{12}\right)=\sqrt{\dfrac{1-\dfrac{\sqrt3}{2}}{2}}$

$\longrightarrow\cos\left(-\dfrac{5\pi}{12}\right)=\dfrac{\sqrt3-1}{2\sqrt2}$

and,

$\longrightarrow\sin\left(-\dfrac{5\pi}{12}\right)=\sqrt{\dfrac{1-\cos\left(-\dfrac{5\pi}{6}\right)}{2}}$

$\longrightarrow\sin\left(-\dfrac{5\pi}{12}\right)=\sqrt{\dfrac{1+\dfrac{\sqrt3}{2}}{2}}$

$\longrightarrow\sin\left(-\dfrac{5\pi}{12}\right)=-\dfrac{\sqrt3+1}{2\sqrt2}$

Hence the equation of our line becomes,

$\displaystyle\longrightarrow x\left(\dfrac{\sqrt3-1}{2\sqrt2}\right)-y\left(\dfrac{\sqrt3+1}{2\sqrt2}\right)=\sqrt2$

$\displaystyle\longrightarrow\underline{\underline{\left(\sqrt3-1\right)x-\left(\sqrt3+1\right)y=4}}$

Hence (A) is the answer.