The equation of a line making an angle 60^{ci | vedclass

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(A) The slope of the given line $x+y-3=0$ is $m_1 = -1$. Let the slope of the required line be $m$. The angle between the two lines is $60^{\circ}$.Us

Vedclass · Accepted Answer

$(1+\sqrt{3})x+(1-\sqrt{3})y-2=0$

Vedclass · Answer

(A) The slope of the given line $x+y-3=0$ is $m_1 = -1$. Let the slope of the required line be $m$. The angle between the two lines is $60^{\circ}$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m m_1} ight|$,we have $	an 60^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} ight|$.$\sqrt{3} = \left| \frac{m+1}{1-m} ight|$.This gives two cases: $\frac{m+1}{1-m} = \sqrt{3}$ or $\frac{m+1}{1-m} = -\sqrt{3}$.Case $1$: $m+1 = \sqrt{3} - \sqrt{3}m$ $\Rightarrow m(1+\sqrt{3}) = \sqrt{3}-1$ $\Rightarrow m = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$.The equation of the line is $y-1 = (2-\sqrt{3})(x-1)$ $\Rightarrow y-1 = (2-\sqrt{3})x - 2 + \sqrt{3}$ $\Rightarrow (2-\sqrt{3})x - y + (1-\sqrt{3}) = 0$.Case $2$: $m+1 = -\sqrt{3} + \sqrt{3}m$ $\Rightarrow m(1-\sqrt{3}) = -\sqrt{3}-1$ $\Rightarrow m = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}$.The equation of the line is $y-1 = (2+\sqrt{3})(x-1)$ $\Rightarrow y-1 = (2+\sqrt{3})x - 2 - \sqrt{3}$ $\Rightarrow (2+\sqrt{3})x - y - (1+\sqrt{3}) = 0$.Comparing with the options,option $A$ is $(1+\sqrt{3})x+(1-\sqrt{3})y-2=0$ which simplifies to $y-1 = \frac{1+\sqrt{3}}{\sqrt{3}-1}(x-1) = (2+\sqrt{3})(x-1)$,matching Case $2$.

The equation of a line making an angle $60^{\circ}$ with the line $x+y-3=0$ and passing through the point $(1,1)$ is

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