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This is one of our many quizzes on parallel lines. Click on this to access more such competitive quizzes on parallel lines.

Parallel Lines Quiz 2

mathematics

 If a line $\sqrt{19}x+9y+28=0$ is parallel to $2\sqrt{19}x+18y-44=0$ then find the distance between them. 

 Distance between two parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ is calculated as $d=\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.

  $\begin{align} & \text{If two lines }ax+by+{{c}_{1}}=0\text{ and }ax+by+{{c}_{2}}=0\text{ are parallel then the distance between them is:} \\ & d=\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\\ & \text{Two parallel lines are }\sqrt{19}x+9y+28=0\text{ and }2\sqrt{19}x+18y-44=0 \\ & \Rightarrow \sqrt{19}x+9y-22=0 \\  & \Rightarrow \dfrac{\left| 28-(-22) \right|}{\sqrt{19+81}} \\  & \Rightarrow \dfrac{\left| 50 \right|}{\sqrt{100}}\Rightarrow 5 \end{align}$

 Find the equation of a line whose x-intercept is $\dfrac{7}{9}$ and is parallel to $36x+14y-21=0$.  

 Convert the given equation in the form of $y=mx+c$ and then put the value of $m$ in the required slope intercept form $y=m(x-d)$.

 $\begin{align} & \text{Let the required equation be “y=m(x-d)”, where x-intercept, }\\ & d=\dfrac{7}{9}\\ & \text{Since }y=m(x-d)\text{ is parallel to }36x+14y-21=0\text{ i.e., } y=\dfrac{-18}{7}x+\dfrac{3}{2} \\ & \Rightarrow m=\dfrac{-18}{7}\,\,\,\,\,\,\,\,\,\left[ \because \text{Parallel lines have equal slopes} \right] \\ & \therefore \text{Required equation }=y=\dfrac{-18}{7}\left( x-\dfrac{7}{9} \right)\\ & \Rightarrow 18x+7y-14=0\end{align}$

 If the distance between two parallel lines $ax+by-30=0$ and $ax+by-35=0$ is given by $\dfrac{5\sqrt{17}}{34}$ where $b=\dfrac{a}{4}$, then find the value of $a$. 

 Put the given values in $d=\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ and solve for $a$. 

  $\begin{align} & \text{If two lines }ax+by+{{c}_{1}}=0\text{ and }ax+by+{{c}_{2}}=0\text{ are parallel then the distance between them is:} \\ & d=\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \\ & \Rightarrow \dfrac{5\sqrt{17}}{34}=\dfrac{\left| -30-(-35) \right|}{\sqrt{{{a}^{2}}+{{\left( \dfrac{a}{4} \right)}^{2}}}} \\  & \Rightarrow \dfrac{5\sqrt{17}}{34}=\dfrac{5\times 4}{\sqrt{17}a} \\  & \Rightarrow a=\dfrac{34\times 4}{\sqrt{17}\times \sqrt{17}}=8  \end{align}$ 

 Find the value of $a$ for which line passing through the points $\left( \dfrac{-1}{10},\dfrac{a}{3} \right)$ and $\left( \dfrac{1}{10},\dfrac{a}{2} \right)$ is parallel to $45x-6y+17=0$.

 First convert the given equation in the form of $y=mx+c$ and then find the slope of other line using formula $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Now, equate the slopes of both the parallel lines to find the value of $a$.

  $\begin{align}  & \text{Convert the given equation in the form of “y=mx+c”, we get: }\\ & 45x-6y+17=0\Rightarrow y=\dfrac{15}{2}x+\dfrac{17}{6}\\ & \Rightarrow m=\dfrac{15}{2} \\ & \text{Since the parallel lines have equal slopes.} \\ & \Rightarrow \dfrac{\dfrac{a}{2}-\dfrac{a}{3}}{\dfrac{1}{10}+\dfrac{1}{10}}=\dfrac{15}{2}\,\,\,\,\,\,\,\,\left[ \because \text{ Slope of a line}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right] \\  & \Rightarrow \dfrac{5a}{6}=\dfrac{15}{2}\Rightarrow a=9 \end{align}$ 

 Find the condition for which the lines ${{\left( r+s \right)}^{2}}x+2tx+\left( r+s \right)\left( r-s-y \right)=0$ and ${{\left( r-s \right)}^{2}}x-2t\left( x-1 \right)-\left( r+s \right)y=0$are said to be parallel, where $r\ne -s$.  

 First convert the given lines in the form of $y=mx+c$ and then equate the slopes of both the parallel lines to find the condition.

  $\begin{align} & \text{Convert the given lines in the form of }y=mx+c \\  & {{\left( r+s \right)}^{2}}x+2tx+\left( r+s \right)\left( r-s-y \right)=0 \\  & \Rightarrow \left( {{(r+s)}^{2}}+2t \right)x+{{r}^{2}}-{{s}^{2}}-(r+s)y=0 \\  & \Rightarrow y=\dfrac{\left( {{(r+s)}^{2}}+2t \right)}{r+s}x+(r-s) ------(1) \\   & {{\left( r-s \right)}^{2}}x-2t\left( x-1 \right)-\left( r+s \right)y=0 \\  & \Rightarrow \left( {{(r-s)}^{2}}-2t \right)x+2t-\left( r+s \right)y=0 \\  & \Rightarrow y=\dfrac{\left( {{(r-s)}^{2}}-2t \right)}{r+s}x+\dfrac{2t}{r+s} --------(2) \\ & \text{Since parallel lines have equal slopes.} \\ & \Rightarrow \dfrac{\left( {{(r+s)}^{2}}+2t \right)}{r+s}=\dfrac{\left( {{(r-s)}^{2}}-2t \right)}{r+s}\Rightarrow {{(r+s)}^{2}}+2t-{{(r-s)}^{2}}+2t=0 \\  & \Rightarrow {{r}^{2}}+{{s}^{2}}+2rs+4t-{{r}^{2}}-{{s}^{2}}+2rs=0 \\  & \Rightarrow rs+t=0 \end{align}$ 