Question

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

Answer 1

Step 1: Determine the values of $\alpha$

The given lines are,

$L_1:2x-y+3=0$.

$$\begin{gathered} L_2:4x-2y+\alpha =0 \\ \Rightarrow L_2:2\left(2x-y+\frac{\alpha }{2}\right)=0 \\ \Rightarrow L_2:2x-y+\frac{\alpha }{2}=0 \end{gathered}$$

It is known that the distance between two lines $ax+by+c_1=0$ and $ax+by+c_2=0$ can be given by $\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}$.

Thus the distance between $L_1$ and $L_2$ is $P_1=\frac{\left|{\displaystyle \frac{\alpha }{2}}-3\right|}{\sqrt{2^2+1^2}}$

$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{5}}=\frac{\left|{\displaystyle \frac{\alpha -6}{2}}\right|}{\sqrt{5}} \\ \Rightarrow \left|\alpha -6\right|=2 \\ \Rightarrow \alpha -6=\pm 2 \\ \Rightarrow \alpha =8,4 \end{gathered}$$

Therefore, the values of $\alpha$ are $8$ and $4$.

Step 2: Determine the values of $\beta$

The given lines are,

$L_1:2x-y+3=0$.

$$\begin{gathered} L_3:6x-3y+\beta =0 \\ \Rightarrow L_3:3\left(2x-y+\frac{\beta }{3}\right)=0 \\ \Rightarrow L_3:2x-y+\frac{\beta }{3}=0 \end{gathered}$$

It is known that the distance between two lines $ax+by+c_1=0$ and $ax+by+c_2=0$ can be given by $\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}$.

Thus the distance between $L_1$ and $L_3$ is $P_2=\frac{\left|{\displaystyle \frac{\beta }{3}}-3\right|}{\sqrt{2^2+1^2}}$

$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{5}}=\frac{\left|{\displaystyle \frac{\beta -9}{3}}\right|}{\sqrt{5}} \\ \Rightarrow \left|\beta -9\right|=3 \\ \Rightarrow \beta -9=\pm 3 \\ \Rightarrow \beta =12,6 \end{gathered}$$

Therefore, the values of $\beta$ are $12$ and $6$.

Thus, the sum of all possible values of $\alpha$ and $\beta$ can be given by, $\left(8+4+12+6\right)=30$.

Hence, the sum of all possible values of $\alpha$ and $\beta$ is $30$.