Question

find distance between the parallel lines 5x-3y-4 equals to 0 and 10x-6y-9 equals to zero​

Answer 1

$\pink{ \boxed{ \boxed{\begin{array}{cc} \bf \underline{ \maltese\:\:Page-\:1:\:\maltese}\\\\ \maltese \bf \: we \: know \: that\: \\ \\ \sf \: if \: \: \: ax + by + c_1 = 0 \: \: \: \: \: and \\ \sf \: \: ax + by + c_2 = 0 \: \: \: are \: two \: parallel \: \\ \sf \: line \: then \: distance \: between \: them \\ \\ \blue{\sf \: d = \frac{ |c_1 - c_2| }{ \sqrt{ {a}^{2} + {b}^{2} } } \: \: unit} \\ \end{array}}}} \\ \\ \pink{ \boxed{\boxed{\begin{array}{cc} \bf \underline{ \maltese\:\:Page-\:2:\:\maltese}\\\\\maltese \bf \: given \: that \\ \\ \bf \implies \: 5x - 3y - 4 = 0 \: \: \: - - - (1) \\ \\ and \\ \\ \blue{ {{\begin{array}{cc}\: \bf \: \: 10x - 6y - 9 = 0 \\ \bf \implies \: 5x - 3y - \frac{9}{2} = 0 \: \: \: - - - (2) \\ \\ \orange{ {{\begin{array}{cc} \odot \:\: \bf eqn.(1) \: \: and \: \: eqn.(2) \: \: are \: prallel \\ \bf \: so \: distance \: \: between \: them \\ \\ \bf \: d = \frac{ | - 4 - ( - \frac{9}{2} )| }{ \sqrt{ {5}^{2} + {( - 3)}^{2} } } \\ \\ = \frac{ | \frac{9}{2} - 4| }{ \sqrt{25 + 9} } \\ \\ = \frac{ | \frac{9 - 8}{2} | }{ \sqrt{34} } \\ \\ = \frac{ | \frac{ 1}{2} | }{ \sqrt{34} } \\ \\ = \frac{1}{2 \sqrt{34} } \: unit \end{array}}}} \end{array}}}} \: \end{array}}}}$

Answer 2

       $\text{\huge{\underline{Solution:}}}$

       $\text{The two given equations are:}$

       $5x - 3y - 4 = 0 \text{ ------ (i)}$

       $10x - 6y - 9 = 0 \text{ ------ (ii)}$

       $\text{Dividing (ii) by 2}$

$\implies 5x - 6y - \dfrac92 = 0 \text{ ------ (iii)}$

       $\text{We know that, the distance between two parallel lines is:}$

       $d = \dfrac{|c_2-c_1|}{\sqrt{a^{2} + b^{2} }}$

       $\text{In our case:}$

       $a = 5$

       $b = -3$

       $c_1 = -4$

       $c_2 = -\dfrac92$

$\implies d = \dfrac{|-\dfrac92-(-4)|}{\sqrt{(5)^{2} + (-3)^{2}}}$

$\implies d = \dfrac{|-\dfrac92+4|}{\sqrt{25 + 9}}$

$\implies d = \dfrac{|\dfrac{-9+8}2|}{\sqrt{34}}$

$\implies d = \dfrac{|\dfrac{-1}2|}{\sqrt{34}}$

$\implies d = \dfrac{\dfrac{1}2}{\sqrt{34}}$

$\implies d = \boxed{\dfrac{1}{2\sqrt{34}} \text{ units}}$