Question

What is the equation of the line midway between the lines $3x-4y+12=0$ and $3x-4y=6$ ?

• A. $3x-4y-9=0$
• B. $3x-4y+9=0$
• C. $3x-4y-3=0$
• D. $3x-4y+3=0$

Answer 1

Answer: (D)

$3x-4y+3=0$

The line which is midway to both the lines should be parallel to them.

Let the line be $3x-4y=c$, the Y-intercept of the required line should be midway of Y-intercept of the other 2 lines

Y-intercept of $1^{st}$ line-

$\Rightarrow 3\left(0\right)-4y=-12$

$\Rightarrow y=3$

Y-intercept of $2^{nd}$ line -

$\Rightarrow 3\left(0\right)-4y=6$

$\Rightarrow y=\frac{-3}{2}$

Y-intercept of required line-

$3\left(0\right)-4y=c$

$\Rightarrow y=-\frac{c}{4}$

$\Rightarrow -\frac{c}{4}=\frac{3+\frac{(-3)}{2}}{2}$

$\Rightarrow c=-3$

The required line is given by $3x-4y+3=0$