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(A) The equation of a line passing through $(1, 0)$ with slope $m$ is given by $y - 0 = m(x - 1)$,which simplifies to $mx - y - m = 0$.The perpendicul

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$\sqrt{3}x + y - \sqrt{3} = 0, \sqrt{3}x - y - \sqrt{3} = 0$

Vedclass · Answer

(A) The equation of a line passing through $(1, 0)$ with slope $m$ is given by $y - 0 = m(x - 1)$,which simplifies to $mx - y - m = 0$.The perpendicular distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$A = m, B = -1, C = -m$,and $d = \frac{\sqrt{3}}{2}$.Substituting these values: $\frac{|-m|}{\sqrt{m^2 + (-1)^2}} = \frac{\sqrt{3}}{2}$.Squaring both sides: $\frac{m^2}{m^2 + 1} = \frac{3}{4}$.$4m^2 = 3m^2 + 3 \implies m^2 = 3 \implies m = \pm \sqrt{3}$.For $m = \sqrt{3}$,the equation is $\sqrt{3}x - y - \sqrt{3} = 0$.For $m = -\sqrt{3}$,the equation is $-\sqrt{3}x - y + \sqrt{3} = 0$,which is $\sqrt{3}x + y - \sqrt{3} = 0$.

The equations of the lines passing through the point $(1, 0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin are:

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