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  # Find the area bounded by the lines $y=4x+5,y=5-x$ and $4y=x+5.$
  
  $\begin{array}{1 1} \frac{13}{1}\;sq.units. \\ 26\;sq.units. \\30\;sq.units. \\ \frac{15}{2}\;sq.units\end{array}$
  
  Toolbox:
  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c. Step 1: Given :y=4x+5,y=5-x and 4y=x+5. First let us sketch the graph of the given lines. Consider y=4x+5 Clearly the lines meets the x-axis at (-\large\frac{5}{4},0) and the y-axis at (0,5).Join these two points,we get the graph of the above line.Similarly the graphs of the other two lines can be obtained. Step 2: Now let us find the points of intersection of these lines. 4x-y=-5-------(1) x+y=5-------(2) _______________(+) 5x=0\Rightarrow x=0 and y=5 x+y=5 x-4y=-5-------(3) ________________(-) 5y=10\Rightarrow y=2 and x=3 \qquad\qquad\;4x-y=-5 (3)\times 4\Rightarrow 4x-16y=-20 ________________________ 15y=15\Rightarrow y=1 and x=-1 Hence the points of intersection are (-1,1),(0,5) and (3,2) Step 3: Now the required area is the shaded portion shown in the fig. Area of the shaded portion is A=\int_{-1}^0y_1dx+\int_0^3y_2dx-\int_{-1}^3y_3dx. On substituting for y_1,y_2 and y_3 A=\int_{-1}^0(4x+5)dx+\int_0^3(5-x)dx-\int_{-1}^3\large\frac{x+5}{4}$$dx.$
  On integrating we get,
  $A=\begin{bmatrix}\large\frac{4x^2}{2}\normalsize dx+5x\end{bmatrix}_{-1}^0+\begin{bmatrix}5x-\large\frac{x^2}{2}\end{bmatrix}_0^3-\large\frac{1}{4}\begin{bmatrix}\large\frac{x^2}{2}+\normalsize 5x\end{bmatrix}_{-1}^3$
  $\;\;=2(x^2)_{-1}^0+5(x)_{-1}^0+5(x)_0^3-\large\frac{1}{2}$$(x^2)_0^3-\large\frac{1}{8}$${(x^2)}_{-1}^3-\large\frac{5}{4}$$(x)_{-1}^3 Step 4: On applying limits, \;\;=2[0-1]+5[0+1]+5[3-0]-\large\frac{1}{2}$$[9-0]-\large\frac{1}{8}$$[9-1]-\large\frac{5}{4}$$[3+1]$
  On simplifying we get,
  $\;\;=-2+5+15-\large\frac{9}{2}$$-1-5$
  $\;\;=12-\large\frac{9}{2}=\large\frac{15}{2}$sq.units.
  edited May 7, 2013
explanation: The document contains a math question about finding the area bounded by three given lines. It also includes a detailed solution to the problem, leading to the final answer of \(\frac{15}{2}\) square units.
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  # Find the area bounded by the lines $y=4x+5,y=5-x$ and $4y=x+5.$
  
  $\begin{array}{1 1} \frac{13}{1}\;sq.units. \\ 26\;sq.units. \\30\;sq.units. \\ \frac{15}{2}\;sq.units\end{array}$
  
  Toolbox:
  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c. Step 1: Given :y=4x+5,y=5-x and 4y=x+5. First let us sketch the graph of the given lines. Consider y=4x+5 Clearly the lines meets the x-axis at (-\large\frac{5}{4},0) and the y-axis at (0,5).Join these two points,we get the graph of the above line.Similarly the graphs of the other two lines can be obtained. Step 2: Now let us find the points of intersection of these lines. 4x-y=-5-------(1) x+y=5-------(2) _______________(+) 5x=0\Rightarrow x=0 and y=5 x+y=5 x-4y=-5-------(3) ________________(-) 5y=10\Rightarrow y=2 and x=3 \qquad\qquad\;4x-y=-5 (3)\times 4\Rightarrow 4x-16y=-20 ________________________ 15y=15\Rightarrow y=1 and x=-1 Hence the points of intersection are (-1,1),(0,5) and (3,2) Step 3: Now the required area is the shaded portion shown in the fig. Area of the shaded portion is A=\int_{-1}^0y_1dx+\int_0^3y_2dx-\int_{-1}^3y_3dx. On substituting for y_1,y_2 and y_3 A=\int_{-1}^0(4x+5)dx+\int_0^3(5-x)dx-\int_{-1}^3\large\frac{x+5}{4}$$dx.$
  On integrating we get,
  $A=\begin{bmatrix}\large\frac{4x^2}{2}\normalsize dx+5x\end{bmatrix}_{-1}^0+\begin{bmatrix}5x-\large\frac{x^2}{2}\end{bmatrix}_0^3-\large\frac{1}{4}\begin{bmatrix}\large\frac{x^2}{2}+\normalsize 5x\end{bmatrix}_{-1}^3$
  $\;\;=2(x^2)_{-1}^0+5(x)_{-1}^0+5(x)_0^3-\large\frac{1}{2}$$(x^2)_0^3-\large\frac{1}{8}$${(x^2)}_{-1}^3-\large\frac{5}{4}$$(x)_{-1}^3 Step 4: On applying limits, \;\;=2[0-1]+5[0+1]+5[3-0]-\large\frac{1}{2}$$[9-0]-\large\frac{1}{8}$$[9-1]-\large\frac{5}{4}$$[3+1]$
  On simplifying we get,
  $\;\;=-2+5+15-\large\frac{9}{2}$$-1-5$
  $\;\;=12-\large\frac{9}{2}=\large\frac{15}{2}$sq.units.
  edited May 7, 2013
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distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question about finding the area bounded by three given lines. It also includes a detailed solution to the problem, leading to the final answer of \(\frac{15}{2}\) square units.
  
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