Question

find the distance between the parallel lines 3x + 2y = 6 and 3x + 2y = 12​

Answer 1

$\pink{ \boxed{ \boxed{\begin{array}{cc} \bf \underline{ \maltese\:\:Page-\:1:\:\maltese}\\\\ \maltese \bf \: we \: know \: that\: \\ \\ \sf \: if \: \: \: ax + by + c_1 = 0 \: \: \: \: \: and \\ \sf \: \: ax + by + c_2 = 0 \: \: \: are \: two \: parallel \: \\ \sf \: line \: then \: distance \: between \: them \\ \\ \blue{\sf \: d = \frac{ |c_1 - c_2| }{ \sqrt{ {a}^{2} + {b}^{2} } } \: \: unit} \\ \end{array}}}}$

$\pink{ \boxed{\boxed{\begin{array}{cc} \bf \underline{ \maltese\:\:Page-\:2:\:\maltese}\\\\\maltese \bf \: given \: that \\ \\ \bf \: 3x + 2y = 6 \\ \bf \implies \: 3x + 2y - 6 = 0 \: \: \: - - - (1) \\ \\ and \\ \\ \blue{ {{\begin{array}{cc}\: \bf \: \: 3x + 2y = 12 \\ \bf \implies \: 3x + 2y - 12 = 0 \: \: \: - - - (2) \\ \\ \orange{ {{\begin{array}{cc} \odot \:\: \bf eqn.(1) \: \: and \: \: eqn.(2) \: \: are \: prallel \\ \bf \: so \: distance \: \: between \: them \\ \\ \bf \: d = \frac{ | - 6 -( - 12)| }{ \sqrt{ {3}^{2} + {2}^{2} } } \\ \\ = \frac{ | - 6 + 12| }{ \sqrt{9 + 4} } \\ \\ = \frac{ | 6| }{ \sqrt{13} } \\ \\ \bf \therefore \: \: d = \frac{6}{ \sqrt{13} } \: \: unit \: \end{array}}}} \end{array}}}} \: \end{array}}}}$