
How do you graph \[y=-\dfrac{5}{4}x-3\]?
Answer
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Hint: We are asked to draw the graph of the equation \[y=-\dfrac{5}{4}x-3\]. 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 \[y=-\dfrac{5}{4}x-3\], the equation can also be written as \[-\dfrac{5}{4}x-y-3=0\], comparing with the general equation of straight line, we get \[a=-\dfrac{5}{4},b=-1\And c=-3\].
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 \[-\dfrac{5}{4}x-y-3=0\], we get
\[\begin{align}
& \Rightarrow -\dfrac{5}{4}x-0-3=0 \\
& \Rightarrow -\dfrac{5}{4}x-3=0 \\
\end{align}\]
Solving the above equation, we get
\[\Rightarrow x=-\dfrac{12}{5}\]
So, the coordinates of the X-intercept are \[\left( -\dfrac{12}{5},0 \right)\].
Similarly, now we substitute \[x=0\] in the equation \[-\dfrac{5}{4}x-y-3=0\], we get
\[\begin{align}
& \Rightarrow -\dfrac{5}{4}(0)-y-3=0 \\
& \Rightarrow -y-3=0 \\
\end{align}\]
Adding \[y\]to both sides of the equation, we get
\[\therefore y=-3\]
So, the coordinates of the Y-intercept are \[\left( 0,-3 \right)\].
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 O either- \[x\] 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 \[y=-\dfrac{5}{4}x-3\], the equation can also be written as \[-\dfrac{5}{4}x-y-3=0\], comparing with the general equation of straight line, we get \[a=-\dfrac{5}{4},b=-1\And c=-3\].
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 \[-\dfrac{5}{4}x-y-3=0\], we get
\[\begin{align}
& \Rightarrow -\dfrac{5}{4}x-0-3=0 \\
& \Rightarrow -\dfrac{5}{4}x-3=0 \\
\end{align}\]
Solving the above equation, we get
\[\Rightarrow x=-\dfrac{12}{5}\]
So, the coordinates of the X-intercept are \[\left( -\dfrac{12}{5},0 \right)\].
Similarly, now we substitute \[x=0\] in the equation \[-\dfrac{5}{4}x-y-3=0\], we get
\[\begin{align}
& \Rightarrow -\dfrac{5}{4}(0)-y-3=0 \\
& \Rightarrow -y-3=0 \\
\end{align}\]
Adding \[y\]to both sides of the equation, we get
\[\therefore y=-3\]
So, the coordinates of the Y-intercept are \[\left( 0,-3 \right)\].
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 O either- \[x\] 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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