The equations of two lines passing through (0 | vedclass

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(C) Let the equation of a line passing through $(0, a)$ be $y - a = m(x - 0)$,which simplifies to $mx - y + a = 0$ ... $(i)$.The perpendicular distanc

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$y - a = 0$ and $4x - 3y + 3a = 0$

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(C) Let the equation of a line passing through $(0, a)$ be $y - a = m(x - 0)$,which simplifies to $mx - y + a = 0$ ... $(i)$.The perpendicular distance from the point $(2a, 2a)$ to the line $(i)$ is given as $a$.Using the distance formula,we have: $a = \frac{|m(2a) - (2a) + a|}{\sqrt{m^2 + 1}}$.$a = \frac{|2am - a|}{\sqrt{m^2 + 1}} \Rightarrow a\sqrt{m^2 + 1} = |a(2m - 1)|$.Since $a 
eq 0$,we have $\sqrt{m^2 + 1} = |2m - 1|$.Squaring both sides: $m^2 + 1 = (2m - 1)^2 \Rightarrow m^2 + 1 = 4m^2 - 4m + 1$.$3m^2 - 4m = 0 \Rightarrow m(3m - 4) = 0$.Thus,$m = 0$ or $m = \frac{4}{3}$.For $m = 0$,the equation is $y - a = 0$.For $m = \frac{4}{3}$,the equation is $y - a = \frac{4}{3}x$ $\Rightarrow 3y - 3a = 4x$ $\Rightarrow 4x - 3y + 3a = 0$.Therefore,the required equations are $y - a = 0$ and $4x - 3y + 3a = 0$.

The equations of two lines passing through $(0, a)$ which are at a distance of $a$ from the point $(2a, 2a)$ are:

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