Question

How do you graph $x - 2y = 3$?

Answer 1

Hint : We will use intercept form to determine the points of the given equation. And, we will also solve this question by assuming the value of $x = 0$ , by applying the value of $x$ , we will get the coordinate of $y$ . Then, we will assume the value of $y = 0$ , by which we will get the $x$ coordinate. Finally, we will plot the points in the graph.

Complete step-by-step answer :
Here, we will graph $x - 2y = 3$.
Now, we will write the equation in the slope- intercept form i.e., $y = mx + b$ $\to \left( 1 \right)$
Where the $m$ is the slope
$b$ is the $y$ - intercept
Then, we have $- 2y = 3 - x$
$y = \dfrac{{3 - x}}{{ - 2}}$
$y = \dfrac{{ - 3}}{2} + \dfrac{x}{2}$
$y = \dfrac{1}{2}x - \dfrac{3}{2}$ $\to \left( 2 \right)$
By comparing equation $\left( 1 \right)$ and $\left( 2 \right)$ , we have
$m = \dfrac{1}{2}$ i.e., the slope of the equation
$b = - \dfrac{3}{2}$ i.e., the $y$ - intercept
The $y$ - intercept is the point where the line intersects the $y$ -axis.
Therefore, the point is $\left( {0, - \dfrac{3}{2}} \right)$ .
Slope is the ‘steepness’ of the line, also commonly known as rise over run i.e., $\dfrac{{rise}}{{run}}$ . Here, $m = \dfrac{1}{2}$ therefore, we can say that the graph “rise” $1$ point upwards and “runs” $2$ points to the right from the $y$ - intercept.
Now, we know the slope and the $y$ - intercept, thus we also know that $\left( {0 + 2, - \dfrac{3}{2} + 1} \right) = \left( {2, - \dfrac{1}{2}} \right)$ which will also be on the line.
Now, we know two points of the equation i.e., $\left( {0, - \dfrac{3}{2}} \right)$ and $\left( {2, - \dfrac{1}{2}} \right)$ .
Let us plot these points graphically,

Note : Equation of straight line is usually written in the slope-intercept form. When we are given an equation in slope- intercept form, we can use the $y$ - intercept as the point, then out the slope from there. When an equation of a line is not given in slope-intercept form, our first step will be to solve the equation for $y$ . Sometimes the slope intercept form will be called as $y$ -form.