Question

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

Answer 1

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
$\begin{aligned} & 3x+2y=4 \\\ & \Rightarrow \dfrac{3x}{4}+\dfrac{2y}{4}=1 \\\ & \Rightarrow \dfrac{x}{{}^{4}/{}_{3}}+\dfrac{y}{2}=1 \\\ \end{aligned}$
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 $\infty$