Question

The equation of the line parallel to the line $2x-3y=1$ and passing through the middle point of the line segment joining the points $(1,3)$ and $(1,-7)$, is:

• A. $2x-3y+8=0$
• B. $2x-3y=8$
• C. $2x-3y+4=0$
• D. $2x-3y=4$

Answer 1

The correct option is B

$2x-3y=8$

Explanation for the correct answer:

Finding the equation of the line,

Let the required line be $AB$.

Given that $AB$ is parallel to $2x-3y=1$.

Therefore, the slope of $AB$ and $2x-3y=1$ will be the same.

Slope of $2x-3y=1$:

$$\begin{gathered} 2x-3y=1 \\ \Rightarrow y=\frac{2}{3}x-1\left[\text{equation is of the form}y=mx+c\right] \\ \therefore \text{slope}=\frac{2}{3} \end{gathered}$$

Given that $AB$ passes through the midpoint of $(1,3)$ and $(1,-7)$:

Calculating the midpoint of $(1,3)$ and $(1,-7)$:
$$\begin{gathered} =\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \\ =\left(\frac{1+1}{2},\frac{3-7}{2}\right) \\ =(1,-2) \end{gathered}$$

Finding the equation of $AB$

$AB$ passes through the point $(1,-2)$ and has slope $\frac{2}{3}$.

$$\begin{gathered} y-y_1=m(x-x_1) \\ y+2=\frac{2}{3}(x-1) \\ \Rightarrow 3y+6-2x+2=0 \\ \Rightarrow 2x-3y=8 \end{gathered}$$

Hence, the correct answer is Option (B).