Question

Find the equation of the straight lines passing through the origin and making an angle of ${45}^{\circ }$ with the straight line $\sqrt{3}x+y=11$

Answer 1

Let the required equation be $ax+by=c$ but here it passes through origin (0, 0)

$\therefore$ Equation is $ax+by=0$

Slope of the line $\left(m_1\right)=\frac{-a}{b}$ and $\left(m_2\right)=\frac{-\sqrt{3}}{l}$

$\Rightarrow$ Angle between $\sqrt{3}x+y=11$ and $ax+by=0$ is ${45}^{\circ }$

$\therefore tan{45}^{\circ }=\frac{m_1\pm m_2}{1\mp m_1{m}_2}$

$1=\frac{\frac{-a}{b}\pm (-\sqrt{3})}{1\mp \frac{a}{b}\sqrt{3}}$

$1-\frac{\sqrt{3a}}{b}=\frac{-a}{b}-\sqrt{3}$ and $1+\frac{a}{b}\sqrt{3}=\frac{-a}{b}\sqrt{3}$

$b-\sqrt{3}a=-a-\sqrt{3}b$ and $b+a\sqrt{3}=-a+b\sqrt{3}$

$a(1-\sqrt{3})=b(-\sqrt{3}-1)$ and $a(\sqrt{3}+1)=b(\sqrt{3}-1)$

$\frac{a}{b}-\frac{1-\sqrt{3}}{\sqrt{3}-1}=\frac{(\sqrt{3}-1{)}^2}{2}=2-\sqrt{3}$

or

$\frac{a}{b}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=-2-\sqrt{3}$

$\therefore$ Required lines are $\frac{y}{x}=\sqrt{3}\pm 2$ or $y=(\sqrt{3}\pm 2)x$