A line L passing through the point (2,0) make | vedclass

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(C) The slope of the given line $2x-y+3=0$ is $m_1 = 2$. Let the slope of line $L$ be $m$. The angle between the lines is $60^{\circ}$,so $	an 60^{\c

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$\frac{16-10 \sqrt{3}}{11}$

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(C) The slope of the given line $2x-y+3=0$ is $m_1 = 2$. Let the slope of line $L$ be $m$. The angle between the lines is $60^{\circ}$,so $	an 60^{\circ} = |\frac{m-m_1}{1+m m_1}|$. $\sqrt{3} = |\frac{m-2}{1+2m}|$. This gives two cases: Case $1$: $\frac{m-2}{1+2m} = \sqrt{3} \Rightarrow m-2 = \sqrt{3} + 2\sqrt{3}m \Rightarrow m(1-2\sqrt{3}) = 2+\sqrt{3} \Rightarrow m = \frac{2+\sqrt{3}}{1-2\sqrt{3}} = \frac{(2+\sqrt{3})(1+2\sqrt{3})}{1-12} = \frac{2+4\sqrt{3}+\sqrt{3}+6}{-11} = \frac{8+5\sqrt{3}}{-11}$. Case $2$: $\frac{m-2}{1+2m} = -\sqrt{3} \Rightarrow m-2 = -\sqrt{3} - 2\sqrt{3}m \Rightarrow m(1+2\sqrt{3}) = 2-\sqrt{3} \Rightarrow m = \frac{2-\sqrt{3}}{1+2\sqrt{3}} = \frac{(2-\sqrt{3})(1-2\sqrt{3})}{1-12} = \frac{2-4\sqrt{3}-\sqrt{3}+6}{-11} = \frac{8-5\sqrt{3}}{-11}$. Since $L$ makes an acute angle with the positive $X$-axis,$m > 0$. Comparing the two slopes,$\frac{8-5\sqrt{3}}{-11} = \frac{5\sqrt{3}-8}{11} \approx \frac{5(1.732)-8}{11} = \frac{8.66-8}{11} > 0$. Thus,$m = \frac{5\sqrt{3}-8}{11}$. The equation of line $L$ is $y - 0 = m(x - 2) \Rightarrow y = \frac{5\sqrt{3}-8}{11}(x - 2)$. The $Y$-intercept is found by setting $x = 0$: $y = \frac{5\sqrt{3}-8}{11}(-2) = \frac{16-10\sqrt{3}}{11}$.

$A$ line $L$ passing through the point $(2,0)$ makes an angle $60^{\circ}$ with the line $2x-y+3=0$. If $L$ makes an acute angle with the positive $X$-axis in the anticlockwise direction,then the $Y$-intercept of the line $L$ is:

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