The equation of line that is parallel to 3x = 2 - 2y and passes through (3, -6) is y = -x - 3
Solution:
Given, line equation is 3x = 2 – 3y
⇒ 3x + 3y = 2 ⇒ (1)
The point is (3, -6)
We have to find the equation of a line which is parallel to given line and passes through given point.
Now, let us find the slope of the given equation
![\text { slope }=\frac{-\mathrm{x} \text { coefficient }}{\mathrm{y} \text { coefficient }}=(-3)/(3)=-1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tzad5e49ee3tpbpy2y7dl1boejsdhlb7rb.png)
So, slope of line is -1
We also know that, slopes of parallel lines are equal, then slope of our required line is – 1.
Then, let us find our line equation using point slope form
![\mathrm{y}-\mathrm{y}_(1)=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_(1)\right)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5tmktqwi1yfmuv2nvsfk7s2ceqtdwoto01.png)
Where "m" is the slope of the line
![\text { Here in our problem, } m=-1,\left(x_(1), y_(1)\right)=(3,-6)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/57on17qa08dj2r2x3xe7ozmy7bviq36i01.png)
By substituting the values in point slope form we get,
![\begin{array}{l}{\rightarrow y-(-6)=-1(x-3)} \\\\ {\rightarrow y+6=-1(x-3)} \\\\ {\rightarrow y=-x-3}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2yg2re04yr6en7qdwa4iz9sarklj23ya2p.png)
Hence the required equation is y = -x - 3