How do you graph the equation $ y=-1.5x+2 $ ?

Hint: Change of form of the given equation will give the x-intercept and y-intercept of the line $ y=-1.5x+2 $ . 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 line to understand the slope and the intercept form of the line $ y=-\dfrac{3}{2}x+2 $ . 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 change from the equation $ y=-1.5x+2 $ to $ 2y=-3x+4 $ .
We have to find the x-intercept, and y-intercept of the line $ 2y=-3x+4 $ .
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 $ 3x+2y=4 $ . Converting into the form of $ \dfrac{x}{p}+\dfrac{y}{q}=1 $ , we get
 $ \begin{align}
  & 3x+2y=4 \\
 & \Rightarrow \dfrac{3x}{4}+\dfrac{2y}{4}=1 \\
 & \Rightarrow \dfrac{x}{{}^{4}/{}_{3}}+\dfrac{y}{2}=1 \\
\end{align} $
Therefore, the x intercept, and y intercept of the line $ y=-1.5x+2 $ is $ \dfrac{4}{3} $ and 2 respectively. The axes intersecting points are $ \left( \dfrac{4}{3},0 \right),\left( 0,2 \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. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to $ \infty $