Q.Show that the lines $2x+3y-6=0$ and $4x+6y+5=0$ are parallel, and find the distance between them.

Two lines are parallel exactly when their $x$ and $y$ coefficients are in the same ratio; once confirmed parallel, the distance between them can be found by scaling both to matching coefficients and applying the parallel-line distance formula.

Step 1 — Parallelism. For $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ to be parallel, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$ (and this ratio must differ from $c_1/c_2$, or the lines would be identical, not merely parallel). Here $a_1=2,b_1=3,c_1=-6$ and $a_2=4,b_2=6,c_2=5$.

Both ratios equal $\dfrac12$, confirming the lines are parallel. Checking they are not identical: $\dfrac{c_1}{c_2} = \dfrac{-6}{5} \ne \dfrac12$, so they are genuinely two distinct parallel lines.

Step 2 — Distance. To use the parallel-line distance formula, both equations must share the same $a,b$ coefficients. Divide the second equation by $2$: $4x+6y+5=0 \Rightarrow 2x+3y+2.5=0$. Now both lines are $2x+3y-6=0$ ($c_1=-6$) and $2x+3y+2.5=0$ ($c_2=2.5$), with $a=2,b=3$.

$d = \dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}} = \dfrac{|-6-2.5|}{\sqrt{4+9}} = \dfrac{8.5}{\sqrt{13}} = \dfrac{17}{2\sqrt{13}}$ …

Applying the parallel-line distance formula directly to the two given equations without first scaling them to matching $a,b$ coefficients is a very common error — the formu …