
How do you graph \[y=\dfrac{1}{2}x-5\] by plotting points?
Answer
552.3k+ views
Hint: We are asked to draw the graph of the equation \[y=\dfrac{1}{2}x-5\]. The degree of an equation is the highest power of the variable present in it. So, as for this equation, the highest power present \[x\]is 1, the degree is also 1. From this, it can be said that this is a linear equation. The graph of a linear equation represents a straight line.
Complete step by step solution:
The general equation of a straight line is \[ax+by+c=0\], where \[a,b,c\]are any real numbers. The given equation is \[y=\dfrac{1}{2}x-5\], the equation can also be written as \[\dfrac{1}{2}x-y-5=0\], comparing with the general equation of straight line, we get \[a=\dfrac{1}{2},b=-1\And c=-5\].
To plot the graph of an equation of the straight line, we should know at least two points, through which the line passes.
To make things simple, let’s take the X-intercept and Y-intercept as the two points. X-intercept is the point where the line crosses X-axis, this means that the Y-coordinate will be \[0\], similarly Y-intercept is the point where the line crosses Y-axis, so X-coordinate will be \[0\]. We will use this property now.
We substitute \[y=0\] in the equation \[\dfrac{1}{2}x-y-5=0\], we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}x-0-5=0 \\
& \Rightarrow \dfrac{1}{2}x-5=0 \\
\end{align}\]
Solving the above equation, we get
\[\Rightarrow x=10\]
So, the coordinates of the X-intercept are \[\left( 10,0 \right)\].
Similarly, now we substitute \[x=0\]in the equation \[\dfrac{1}{2}x-y-5=0\], we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}(0)-y-5=0 \\
& \Rightarrow -y-5=0 \\
\end{align}\]
Adding \[y\]to both sides of the equation, we get
\[\begin{align}
& \Rightarrow -y-5+y=y \\
& \therefore y=-5 \\
\end{align}\]
So, the coordinates of the Y-intercept are \[(0,-5)\].
Using these two points we can plot the graph of the equation as follows:
Note:
Here, we found the two points which are X-intercept and Y-intercept by substituting either-or \[y\], one at a time. We can also find these values by converting the straight-line equation to the equation in intercept form which is, \[\dfrac{x}{a}+\dfrac{y}{b}=1\]. Here, \[a\] and \[b\] are X-intercept and Y-intercept respectively.
Complete step by step solution:
The general equation of a straight line is \[ax+by+c=0\], where \[a,b,c\]are any real numbers. The given equation is \[y=\dfrac{1}{2}x-5\], the equation can also be written as \[\dfrac{1}{2}x-y-5=0\], comparing with the general equation of straight line, we get \[a=\dfrac{1}{2},b=-1\And c=-5\].
To plot the graph of an equation of the straight line, we should know at least two points, through which the line passes.
To make things simple, let’s take the X-intercept and Y-intercept as the two points. X-intercept is the point where the line crosses X-axis, this means that the Y-coordinate will be \[0\], similarly Y-intercept is the point where the line crosses Y-axis, so X-coordinate will be \[0\]. We will use this property now.
We substitute \[y=0\] in the equation \[\dfrac{1}{2}x-y-5=0\], we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}x-0-5=0 \\
& \Rightarrow \dfrac{1}{2}x-5=0 \\
\end{align}\]
Solving the above equation, we get
\[\Rightarrow x=10\]
So, the coordinates of the X-intercept are \[\left( 10,0 \right)\].
Similarly, now we substitute \[x=0\]in the equation \[\dfrac{1}{2}x-y-5=0\], we get
\[\begin{align}
& \Rightarrow \dfrac{1}{2}(0)-y-5=0 \\
& \Rightarrow -y-5=0 \\
\end{align}\]
Adding \[y\]to both sides of the equation, we get
\[\begin{align}
& \Rightarrow -y-5+y=y \\
& \therefore y=-5 \\
\end{align}\]
So, the coordinates of the Y-intercept are \[(0,-5)\].
Using these two points we can plot the graph of the equation as follows:
Note:
Here, we found the two points which are X-intercept and Y-intercept by substituting either-or \[y\], one at a time. We can also find these values by converting the straight-line equation to the equation in intercept form which is, \[\dfrac{x}{a}+\dfrac{y}{b}=1\]. Here, \[a\] and \[b\] are X-intercept and Y-intercept respectively.
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