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What is the equation of the line mid-way between the lines $3 x-4 y+12=0$ and $3 x-4 y=6$ ?
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The correct answer is:
$3 x-4 y+3=0$
$3 x-4 y+12=0$ or $y=\frac{3}{4} x+3$
$3 x-4 y=6 \quad$ or $\quad y=\frac{3}{4} x-\frac{3}{2}$
Equation of line mid-way between these two lines $\mathrm{y}=\frac{3}{4} \mathrm{x}+\left(\frac{3-\frac{3}{2}}{2}\right)$
$\mathrm{y}=\frac{3}{4} \mathrm{x}+\frac{3}{4}$
$4 y=3 x+3$
$3 x-4 y+3=0$
$3 x-4 y=6 \quad$ or $\quad y=\frac{3}{4} x-\frac{3}{2}$
Equation of line mid-way between these two lines $\mathrm{y}=\frac{3}{4} \mathrm{x}+\left(\frac{3-\frac{3}{2}}{2}\right)$
$\mathrm{y}=\frac{3}{4} \mathrm{x}+\frac{3}{4}$
$4 y=3 x+3$
$3 x-4 y+3=0$
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