Question

If the lines $3\left(\alpha -1\right)y-6x=2$ and $4y-8x+10=0$ are parallel, then find the value of $\alpha$.

Answer 1

Step 1:Finding the slope of $3\left(\alpha -1\right)y-6x=2$

$$\begin{gathered} 3\left(\alpha -1\right)y-6x=2 \\ \Rightarrow y=\frac{6x+2}{3\left(\alpha -1\right)} \\ \Rightarrow y=\frac{6}{3\left(\alpha -1\right)}x+\frac{2}{3\left(\alpha -1\right)} \end{gathered}$$

On comparing with the general equation of a line $y=m_{1}x+c$

$\Rightarrow m_1=\frac{6}{3\left(\alpha -1\right)}..........................\left(1\right)$

Step 2:Finding the slope of $4y-8x+10=0$

$$\begin{gathered} 4y-8x+10=0 \\ \Rightarrow 4y=8x-10 \\ \Rightarrow y=2x-\frac{10}{4} \end{gathered}$$

In comparing with the general equation of a line $y=m_{2}x+c$

$\Rightarrow m_2=2................................................\left(2\right)$

Step 3:Finding the value of $\alpha$:

Since both lines are parallel to each other, therefore, the slope of both lines will be the same

From equations (1) and (2)

$\begin{array}{rcl}{m}_1& =& m_2\\ & \Rightarrow & \frac{6}{3\left(\alpha -1\right)}=2\\ & \Rightarrow & 6=2\times 3\left(\alpha -1\right)\\ & \Rightarrow & \alpha -1=1\\ & \Rightarrow & \alpha =2\end{array}$

Therefore, the value of $\alpha$ is $2$.