If the line, 2x - y + 3 = 0 is at a distance 1/spart5 and

Question:

If the line, $2 x-y+3=0$ is at a distance $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$ from the lines $4 x-2 y+\alpha=0$ and $6 x-3 y+\beta=0$,

respectively, then the sum of all possible values of $\alpha$ and $\beta$ is________

Solution:

Apply distance between parallel line formula

$4 x-2 y+\alpha=0$

$4 x-2 y+6=0$

$\left|\frac{\alpha-6}{255}\right|=\frac{1}{55}$

$|\alpha-6|=2 \Rightarrow \alpha=8,4$

sum $=12$

again

$6 x-3 y+\beta=0$

$6 x-3 y+9=0$

$\left|\frac{\beta-9}{3 \sqrt{5}}\right|=\frac{2}{\sqrt{5}}$

$|\beta-9|=6 \Rightarrow \beta=15,3$

sum $=18$

sum of all values of $\alpha$ and $\beta$ is $=30$

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