Question

The distance between $4x+3y=11$ and $8x+6y=15$ is:

• A. $\frac{7}{2}$
• B. $4$
• C. $\frac{7}{10}$
• D. None of these

Answer 1

The correct option is C

$\frac{7}{10}$

Step 1 : Finding the slope of the two lines.

The given equations are: $4x+3y=11$ and $8x+6y=15$

Slope of $4x+3y=11$:

$$\begin{gathered} 4x+3y=11 \\ \Rightarrow y=\frac{-4}{3}x+11\left[\text{equation is of the form}y=mx+c\right] \\ \mathrm{slope}=-\frac{4}{3} \end{gathered}$$

Slope of $8x+6y=15$:

$$\begin{gathered} 8x+6y=15 \\ \Rightarrow y=-\frac{4}{3}x+\frac{15}{2}\left[\text{equation is of the form}y=mx+c\right] \\ \text{slope}=-\frac{4}{3} \end{gathered}$$

Step 2 : Finding the distance between the two lines.

As the slope of the two equations are the same, the lines are parallel to each other.

We see that: $c_1=11,c_2=15,a=-4,b=3$

Therefore, the distance between the two lines is

$$\begin{gathered} \frac{c_1-c_2}{\sqrt{a^2+b^2}} \\ =\frac{11-{\displaystyle \frac{15}{2}}}{\sqrt{16+9}} \\ =\frac{7}{10} \end{gathered}$$

Hence, the correct answer is Option (C).