Question

Find the distance between the parallel lines 3x+4y-3=0 and 6x+8y-1=0

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Parallel lines are}$

$\mathsf{3x+4y-3=0\;\;and \;\;6x+8y-1=0}$

$\underline{\textbf{To find:}}$

$\textsf{The distance between the parallel lines}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{3x+4y-3=0\;\;and \;\;6x+8y-1=0}$

$\textsf{Divide bothsides of the second equation by 2}$

$\mathsf{3x+4y-3=0\;\;and \;\;3x+4y-\dfrac{1}{2}=0}$

$\textbf{Distance between the parallel lines}$

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

$\mathsf{=\left|\dfrac{-3-\left(\dfrac{-1}{2}\right)}{\sqrt{3^2+4^2}}\right|}$

$\mathsf{=\left|\dfrac{-3+\dfrac{1}{2}}{\sqrt{9+16}}\right|}$

$\mathsf{=\left|\dfrac{\dfrac{-6+1}{2}}{\sqrt{25}}\right|}$

$\mathsf{=\left|\dfrac{\dfrac{-5}{2}}{5}\right|}$

$\mathsf{=\left|\dfrac{-1}{2}\right|}$

$\mathsf{=\dfrac{1}{2}}$

$\therefore\textbf{The distance between the given parallel lines}$

$\bf\dfrac{1}{2}\;\textbf{units}$

$\underline{\textbf{Formula used:}}$

$\boxed{\begin{minipage}{6cm} \\\mathsf{The\;distance\;between\;the\;parallel\;lines}\\\\\mathsf{ax+by+c_1=0\;and\;ax+by+c_2=0\;is}\\\\\mathsf{\left|\dfrac{c_1-c_2}{\sqrt{a^2+b^2}}\right|}\\ \end{minipage}}$

Answer 2

Answer:

Step-by-step explanation: