Question

How do you graph $ y=\dfrac{3}{2}x+3 $ ?

Answer 1

Hint: Change of form of the given equation will give the x-intercept and y-intercept of the line $ y=\dfrac{3}{2}x+3 $. We change it to the form of $ \dfrac{x}{p}+\dfrac{y}{q}=1 $ to find the x-intercept, and y-intercept of the line as $ p $ and $ q $ respectively. then we place the points on the axes and from there we draw the line on the graph.

Complete step by step answer:
We are taking the general equation of a line to understand the slope and the intercept form of the line $ y=\dfrac{3}{2}x+3 $. The given equation is in the form of $ y=mx+k $. m is the slope of the line. The slope of the line is $ \dfrac{3}{2} $ .
We have to find the x-intercept, and y-intercept of the line $ y=\dfrac{3}{2}x+3 $.
For this we convert the given equation into the form of $ \dfrac{x}{p}+\dfrac{y}{q}=1 $ . From the form we get that the x intercept, and y intercept of the line will be $ p $ and $ q $ respectively. The points will be $ \left( p,0 \right),\left( 0,q \right) $ .
The given equation is $ y=\dfrac{3}{2}x+3 $ . Converting into the form of $ \dfrac{x}{p}+\dfrac{y}{q}=1 $ , we get
 $ \begin{align}
  & y=\dfrac{3}{2}x+3 \\
 & \Rightarrow y-\dfrac{3}{2}x=3 \\
 & \Rightarrow \dfrac{x}{-2}+\dfrac{y}{3}=1 \\
\end{align} $
Therefore, the x intercept, and y intercept of the line $ 2x+y=5 $ is 2 and 3 respectively. The axes intercepting points are $ \left( -2,0 \right),\left( 0,3 \right) $ .

Note:
A line parallel to the X-axis does not intersect the X-axis at any finite distance. Hence, we cannot get any finite x-intercept of such a line. The same goes for lines parallel to the Y-axis. In the case of the slope of a line, the range of the slope is 0 to $ \infty $.