91Ó°ÊÓ

The equation of the given line is $4x+3y=6$

The given point is $(0,-3)$

To solve this problem, we will make use of the fact that the slopes of the parallel lines are the same.

• The slope-intercept form of the equation of a line is $y=mx+b$where $m$is the slope and $b$is the y-intercept.
• The slope-point form of an equation of a line with the slope $m$, containing the point$\left(x_1,y_1\right)$is$y-y_1=m(x-x_1)$

The equation of a given line is $4x+3y=6$

Rewriting this equation, we have

$$\begin{gathered} 4x+3y=6 \\ 3y=-4x+6 \\ y=-\frac{4}{3}x+\frac{6}{3} \\ y=-\frac{4}{3}x+2 \end{gathered}$$

This equation is in the slope intercept form. Therefore the slope of the given line is $m=-\frac{4}{3}$

Since the parallel lines have the same slope, we can conclude that the slope of the parallel line is$-\frac{4}{3}$as well.

The point on the parallel line is $(0,-3)$and its slope is $-\frac{4}{3}$

Using the slope-point form we have,

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-(-3))=-\frac{4}{3}(x-0) \\ y+3=-\frac{4}{3}x \\ y=-\frac{4}{3}x-3 \end{gathered}$$

Therefore the equation of the line in the slope-intercept form is$y=-\frac{4}{3}x-3$