Answer
Verified
478.2k+ views
Hint: A line can be completely determined by two points. So, to draw the line just take two points which satisfy the line equation and draw a line through those points in the graph. Plot these lines on the same graph and find the coordinates of their point of intersection. This will give the solution of the given system of equations.
Complete step-by-step answer:
For equation $4x-3y+4=0$
When x = 0 we have $4\times 0-3y+4=0$
$\begin{align}
& \Rightarrow -3y+4=0 \\
& \Rightarrow -3y=0-4 \\
& \Rightarrow -3y=-4 \\
& \Rightarrow y=\dfrac{-4}{-3} \\
\end{align}$
i.e. $y=\dfrac{4}{3}$
When y = 0 we have $4x-3\times 0+4=0$
$\begin{align}
& \Rightarrow 4x-0+4=0 \\
& \Rightarrow 4x+4=0 \\
& \Rightarrow 4x=-4 \\
& \Rightarrow x=\dfrac{-4}{4} \\
\end{align}$
i.e. $x=-1$
We plot the points A $\left( 0,\dfrac{4}{3} \right)$ and B$\left( -1,0 \right)$ and draw a line through the points.
So, we have AB $\equiv 4x-3y+4=0$
For equation $4x+3y-20=0$
When x = 0, $4\times 0+3y-20=0$
$\begin{align}
& \Rightarrow 0+3y-20=0 \\
& \Rightarrow 3y-20=0 \\
& \Rightarrow 3y=20 \\
& \Rightarrow y=\dfrac{20}{3} \\
\end{align}$
i.e. $y=\dfrac{20}{3}$
When y = 0, we have $4x+3\times 0-20=0$
$\begin{align}
& \Rightarrow 4x+0-20=0 \\
& \Rightarrow 4x=20 \\
& \Rightarrow 4x=20 \\
& \Rightarrow x=\dfrac{20}{4} \\
\end{align}$
i.e. $x=5$
We plot points C $\left( 0,\dfrac{20}{3} \right)$ and D \[\left( 5,0 \right)\] and draw line CD on the same graph.
So, CD $\equiv 4x+3y-20=0$
These lines intersect at point E.
Since point E is satisfied by the equations of both the lines, E is the solution of the given system of equations.
As is evident from the graph, area bounded by the lines and x-axis is the area of $\Delta \text{EBD}$
Here E (2,4), B (-1,0) and D (5,0).
We know that area of $\Delta =\dfrac{1}{2}base\times height$
Length of altitude from E to base BD = ordinate of E = 4
Length of BD = \[5-\left( -1 \right)\text{ }=\text{ }5+1\text{ }=\text{ }6\]
Using the formula of area of triangle in $\Delta \text{EBD}$ we get
Hence area of $\Delta \text{EBD}$ = $\dfrac{1}{2}BD\times y-\text{coordinate of E}$ = $\dfrac{1}{2}\times 6\times 4$=$\dfrac{24}{2}=12$
Note: Equation of line in intercept form is
$\dfrac{x}{a}+\dfrac{y}{b}=1$
This line passes through (a, 0) and (0, b)
So, we can directly find these points by writing the equation in intercept form.
e.g. $4x-3y+4=0$
$\begin{align}
& \Rightarrow 4x-3y=-4 \\
& \Rightarrow \dfrac{4x}{-4}-\dfrac{3y}{-4}=1 \\
\end{align}$
i.e. $\dfrac{x}{-1}+\dfrac{y}{\dfrac{4}{3}}=1$
Hence a = -1 and b = $\dfrac{4}{3}$
Hence the line passes through (-1,0) and $\left( 0,\dfrac{4}{3} \right)$
Complete step-by-step answer:
For equation $4x-3y+4=0$
When x = 0 we have $4\times 0-3y+4=0$
$\begin{align}
& \Rightarrow -3y+4=0 \\
& \Rightarrow -3y=0-4 \\
& \Rightarrow -3y=-4 \\
& \Rightarrow y=\dfrac{-4}{-3} \\
\end{align}$
i.e. $y=\dfrac{4}{3}$
When y = 0 we have $4x-3\times 0+4=0$
$\begin{align}
& \Rightarrow 4x-0+4=0 \\
& \Rightarrow 4x+4=0 \\
& \Rightarrow 4x=-4 \\
& \Rightarrow x=\dfrac{-4}{4} \\
\end{align}$
i.e. $x=-1$
x | y |
0 | $\dfrac{4}{3}$ |
-1 | 0 |
We plot the points A $\left( 0,\dfrac{4}{3} \right)$ and B$\left( -1,0 \right)$ and draw a line through the points.
So, we have AB $\equiv 4x-3y+4=0$
For equation $4x+3y-20=0$
When x = 0, $4\times 0+3y-20=0$
$\begin{align}
& \Rightarrow 0+3y-20=0 \\
& \Rightarrow 3y-20=0 \\
& \Rightarrow 3y=20 \\
& \Rightarrow y=\dfrac{20}{3} \\
\end{align}$
i.e. $y=\dfrac{20}{3}$
When y = 0, we have $4x+3\times 0-20=0$
$\begin{align}
& \Rightarrow 4x+0-20=0 \\
& \Rightarrow 4x=20 \\
& \Rightarrow 4x=20 \\
& \Rightarrow x=\dfrac{20}{4} \\
\end{align}$
i.e. $x=5$
x | y |
0 | $\dfrac{20}{3}$ |
5 | 0 |
We plot points C $\left( 0,\dfrac{20}{3} \right)$ and D \[\left( 5,0 \right)\] and draw line CD on the same graph.
So, CD $\equiv 4x+3y-20=0$
These lines intersect at point E.
Since point E is satisfied by the equations of both the lines, E is the solution of the given system of equations.
As is evident from the graph, area bounded by the lines and x-axis is the area of $\Delta \text{EBD}$
Here E (2,4), B (-1,0) and D (5,0).
We know that area of $\Delta =\dfrac{1}{2}base\times height$
Length of altitude from E to base BD = ordinate of E = 4
Length of BD = \[5-\left( -1 \right)\text{ }=\text{ }5+1\text{ }=\text{ }6\]
Using the formula of area of triangle in $\Delta \text{EBD}$ we get
Hence area of $\Delta \text{EBD}$ = $\dfrac{1}{2}BD\times y-\text{coordinate of E}$ = $\dfrac{1}{2}\times 6\times 4$=$\dfrac{24}{2}=12$
Note: Equation of line in intercept form is
$\dfrac{x}{a}+\dfrac{y}{b}=1$
This line passes through (a, 0) and (0, b)
So, we can directly find these points by writing the equation in intercept form.
e.g. $4x-3y+4=0$
$\begin{align}
& \Rightarrow 4x-3y=-4 \\
& \Rightarrow \dfrac{4x}{-4}-\dfrac{3y}{-4}=1 \\
\end{align}$
i.e. $\dfrac{x}{-1}+\dfrac{y}{\dfrac{4}{3}}=1$
Hence a = -1 and b = $\dfrac{4}{3}$
Hence the line passes through (-1,0) and $\left( 0,\dfrac{4}{3} \right)$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE