Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How do you graph $y=\dfrac{1}{2}x-5$ by plotting points?

Last updated date: 24th Jul 2024
Total views: 383.4k
Views today: 7.83k
Verified
383.4k+ 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.