We are told that the line passes through (3,10) and is parallel to y=-9, so, we have something like this:
So, we need to find the yellow line equation.
Standard form:
A straight line equation in standard form is something like this equation:
![Ax+By=C](https://img.qammunity.org/2023/formulas/mathematics/high-school/75j0pzqy8f6gtpgzjampxw030qc85p0hp7.png)
To do this, we can start for here:
![\frac{(y-y_1)}{(x-x_1)_{}}=\text{ m}](https://img.qammunity.org/2023/formulas/mathematics/college/x9niicf9xp9g0b6wky3odmwq6wt2j7cnj6.png)
Where (x_1, y_1) is a point through which the line passes and m is the slope of the line.
In our case the point is (3,10) and the slope is 0 (since the line is parallel to y=-9):
![\begin{gathered} \frac{(y-10)}{(x-3)_{}}=0 \\ y-10=0 \\ y=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/15wu4prhjueodbtm16cp5kj0862sy4cexb.png)
Slope-intercept form:
This equation is something like this:
![y=mx+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/smsb8cbft03lwblmi49nf2l6jby2ofxzws.png)
Where m is, once again, the slope, and b is the y-intercept (the point where the line cross the y axis).
So, we have:
![\begin{gathered} y=0\cdot x+10 \\ y=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bbh5uivnlpog136lt5wtctkghaphjcxj7u.png)
Answer
![\begin{gathered} \frac{(y-10)}{(x-3)_{}}=(0)/(1)\text{ } \\ \\ \text{(y-}10)=0\cdot(x-3) \\ y-10=0\cdot x-0\cdot3 \\ -0x+y=10 \\ 0x+y=10 \\ \\ \text{and} \\ \\ y=0\cdot x+10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3gzaer1fsu7ps3howrv2r67qkfh3dhr2oi.png)