To solve the exercise you can first find the slope of the line using this formula
![\begin{gathered} m=(y_(2)-y_(1))/(x_(2)-x_(1))\Rightarrow\text{ Slope formula} \\ \text{ Where m is the slope and} \\ (x_1,y_1)\text{ and }(x_2,y_2)\text{ are two points through which the line passes} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i8lpyg5tx8xat9t8wvgjkkxo7660i0xe30.png)
In this case, for example, you can take the points (-3,2) and (0,0). So, you have
![\begin{gathered} (x_1,y_1)=(-3,2) \\ (x_2,y_2)=(0,0) \\ m=(0-2)/(0-(-3)) \\ m=(0-2)/(0+3) \\ m=-(2)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ozzjj7z2uyt9f185gsz57m2c0mj7n051qp.png)
Now, you can use the point-slope formula and solve for y. The point-slope formula is
![y-y_1=m(x-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/csobd57zth7rh9k4hz9amldzpq2owf0z4j.png)
So, you have
![\begin{gathered} y-2=-(2)/(3)(x-(-3)) \\ y-2=-(2)/(3)(x+3) \\ \text{ Apply the distributive property to the right side of the equation} \\ y-2=-(2)/(3)\cdot x-(2)/(3)\cdot3 \\ y-2=-(2)/(3)x-2 \\ \text{ Add 2 from both sides of the equation} \\ y-2+2=-(2)/(3)x-2+2 \\ y=-(2)/(3)x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/a56o227i57skeixkrqxepusoi1i030xuin.png)
Therefore, the equation that best represents the graph is
![c\colon y=-(2)/(3)x](https://img.qammunity.org/2023/formulas/mathematics/college/v04729718ao5xhvoe8jz5j4pcwv045kbbo.png)