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This is one of our many quizzes on graphing linear equations. Click on this to access more such competitive quizzes on linear equations.

Graphing linear equations Quiz 2

mathematics

 Study the below given graph and find the equation of a line $g$.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e89dbe26-643e-4cff-bf9a-4c7818dcb0757098125707990759408.png"/>

 Find the slope of the given equation and then find the slope of the required equation. Now, use given points and slope to find the required equation.

  $\text{Given: }3x-2y-4=0\\\Rightarrow y=\dfrac{3}{2}x-2\text{ where slope }= \dfrac{3}{2} \\ \text{Since the line “g” is perpendicular to line “h”.} \\ \Rightarrow \text{Slope of line g}  \times \text{Slope of line h} = -1\\\Rightarrow \text{Slope of line g}  ({{m}_{1}}) = \dfrac{-2}{3}\\\therefore \text{ Required equation of line “g” is} \\ \Rightarrow y-{{y}_{1}}={{m}_{1}}\left( x-{{x}_{1}} \right) \\ \Rightarrow \left( y-\dfrac{19}{13} \right)=\dfrac{-2}{3}\left( x-\dfrac{30}{13} \right) \\ \Rightarrow \dfrac{3\left( 13y-19 \right)}{13}=\dfrac{-2\left( 13x-30 \right)}{13} \\ \Rightarrow 2x+3y-9=0 \\ $ 

 Find the value of $x$ in the below given graph.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/96d47267-c182-4b68-a885-e01441ef6f434380337907620022267.png"/>

 Equate slope of PQ and slope of QR to find the value of $x$.

  $\text{Since the points P, Q and R are collinear.} \\ \text{Slope of PQ = Slope of QR} \\ \Rightarrow \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{{{y}_{3}}-{{y}_{2}}}{{{x}_{3}}-{{x}_{2}}} \\ \Rightarrow \dfrac{-4+5}{2+3}=\dfrac{-3+4}{x-2} \\ \Rightarrow x=7 \\ $ 

 Which of the following graphs best represents the two equations given by $10x+11y+21=0$ and $20x+22y+42=0$?

 Check the ratios of the coefficients of both the equations to find the graphical representation of the lines. 

 On comparing the ratios of the coefficients of $10x+11y+21=0$ and $20x+22y+42=0$, we have: $\\ \dfrac{a}{d}=\dfrac{10}{20}=\dfrac{b}{e}=\dfrac{11}{22}=\dfrac{c}{f}=\dfrac{21}{42} \\$ So, the ratios are equal which implies that both the lines will coincide. Thus, a single line graph is the best graphical representation of both lines.

 Find the price of one packet of gems if 2 packets of gems and 3 chips is Rs. 40 and 3 packets of gems and two chips is Rs. 35.

 Assume the price of one packet of gems and chips be $x$ and $y$. Then form the two linear equations in terms of $x$ and $y$. Now, eliminate $y$ to solve for $x$.

 Let the price of one packet of gems and chips be Rs. $x$ and $y$ respectively. $\\ \Rightarrow 2x+3y=40..........\times 2 \\ \Rightarrow 3x+2y=35..........\times 3 \\ \Rightarrow 4x+6y-9x-6y=80-105 \\ \Rightarrow x=5 \\ $ 

 Find the area of the figure enclosed by the four lines $x=2,x=6,x+2y-8=0$ and the x-axis. 

 First draw the graph using the given four lines. Then find the area using integration.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/03af747d-cb20-465c-ade0-74a9132dd8453145537121394477605.png"/>$\\ \text{Required area of ABCD is} \\ \Rightarrow \int\limits_{2}^{6}{y}\text{ }dx \\ \Rightarrow \int\limits_{2}^{6}{\dfrac{8-x}{2}}\text{ }dx \\ \Rightarrow \dfrac{1}{2}\text{ }\left[ 8x-\dfrac{{{x}^{2}}}{2} \right]_{2}^{6} \\ \Rightarrow \dfrac{1}{2}\text{ }\left[ 48-18-16+2 \right] \\ \Rightarrow 8\text{ sq units} \\ $ 