Question

What is the distance between lines 3x − 5y + 3=0 and 6x − 10y − 12=0?

Answer 1

Let us rewrite the two lines in slope intercept form to obtain,

$3x-5y+3=0$

$\Rightarrow -5y=-3x-3$

$\Rightarrow y=\frac{3}{5}x+\frac{3}{5}$

For the second equation too, we have,

$6x-10y-12=0$

$\Rightarrow -10y=-6x+12$

$\Rightarrow y=\frac{6}{10}x-\frac{12}{10}$

$\Rightarrow y=\frac{3}{5}x-\frac{6}{5}$

The two slopes are equal. this means the two lines are parallel. We apply the formula for finding the distance between 2 parallel lines. This formula is given by

$d=\frac{|b_2-b_1|}{\sqrt{m^2+1} }$

Where

$b_1=\frac{3}{5}$

$b_2=-\frac{6}{5}$

and

$m=\frac{3}{5}$

We substitute into the formula to obtain,

$d=\frac{|-\frac{6}{5}-\frac{3}{5}|}{\sqrt{(\frac{3}{5})^2 +1 } }$

$d=\frac{|\frac{-6-3}{5}|}{\sqrt{\frac{9}{25} +1}}$

$d=\frac{|\frac{-9}{5} |}{\sqrt{\frac{34}{25} } }$

$d=\frac{9\sqrt{34} }{34}$

Therefore the distance between the two lines is approximately 1.5 units.