If the line 2x-y+3=0 is at a distance 1√5 and 2√5 from the lines 4x-2y+=0 and 6x-3y+=0, re…

Question

If the line $2x-y+3=0$ is at a distance $\dfrac1{\sqrt5}$ and $\dfrac2{\sqrt5}$ from the lines $4x-2y+\alpha=0$ and $6x-3y+\beta=0$, respectively, then the sum of all possible value of $\alpha$ and $\beta$ is \_\_\_\_\_\_\_.

Answer: 30

Accepted Answer

$4x-2y+\alpha=0 \equiv 2x-y+\dfrac{\alpha}2=0$. Distance between this and $2x-y+3=0$:
$$\frac{\left|\frac{\alpha}2-3\right|}{\sqrt5}=\frac1{\sqrt5} \implies |\alpha-6|=2 \implies \alpha=4,8$$

$6x-3y+\beta=0 \equiv 2x-y+\dfrac{\beta}3=0$. Distance:
$$\frac{\left|\frac{\beta}3-3\right|}{\sqrt5}=\frac2{\sqrt5} \implies |\beta-9|=6 \implies \beta=15,3$$

$$\text{Sum} = 4+8+15+3=30$$

Solutions · 1