Question

find the distance between the parallel lines 2x-3y+9=0 and 4x-6y+1=0.​

Answer 1

formula:(c1- c2) /(√a1square+ √a2square

2(2x-3y+9=0)

4x-6y+1=0

(9-18)/√16+√36

-9/√52
the distance is in positive so
(9)/2√26

Answer 2

$\mathsf{\Large {\underline {Straight Lines}}}$ :

↪️ 2x - 3y + 9 = 0

4x - 6y + 1 = 0

↪️ 2x - 3y + $\dfrac{1}{2}$ = 0

By Using the Formula,

$\boxed{\mathsf {d\:=\:{\dfrac{|\:C_1\:-\:C_2\:|} {\sqrt{{A} ^{2}\:+\:{B}^{2}}}}}}$

Here, $\mathsf{C_1\:=\:9 \: and \:C_2\:=\:{\dfrac{1}{2}}}$

A = 2, B = 3

Putting the value's in above formula,

$\mathsf {d\:=\:{\dfrac{|\:9\:-\:{\dfrac{1}{2}}\:|} {\sqrt{{2} ^{2}\:+\:{3}^{2}}}}}$

$\mathsf {d\:=\:{\dfrac{|\:{\dfrac{17}{2}}\:|}{{\sqrt{4\:+\:9}}}}}$

$\mathsf {d\:=\:{\dfrac{17}{2\:{\sqrt{13}}}}}$

➡️ $\boxed{\mathsf {d\:=\:{\dfrac{17}{2\:{\sqrt{13}}}}}}$