The general equation of a circle is given as
![\begin{gathered} y=mx+c \\ \text{Where,} \\ m=\text{slope} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uxlssoa1rzc3o8lc7y1pxersj4sddnso4q.png)
The first equation is given as
![2x+y=10](https://img.qammunity.org/2023/formulas/mathematics/college/5wwojookj3d79awx5iq22x3rsbvis3d0yi.png)
Making y the subject of the formula and then comparing coefficient
Subtract 2x from both sides
![\begin{gathered} 2x+y=10 \\ 2x-2x+y=10-2x \\ y=-2x+10 \\ \text{slope}=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/37kxupo5wnnd1nnio772n9awq10xnazftm.png)
The second equation is given as
![-2x+y=8](https://img.qammunity.org/2023/formulas/mathematics/high-school/e1ukgau4o392ae5e2shebhggyyk6fjw5oo.png)
Add 2x to both sides and compare coefficients
![\begin{gathered} -2x+y=8 \\ -2x+2x+y=8+2x \\ y=2x+8 \\ \text{slope}=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ytlowee7b33sjoxvlpyubh9p5fimob4q3f.png)
The third equation is given as
![\begin{gathered} x=-2 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d3s9lrjjiji5pdjbyyjr418rsojvlrtw8d.png)
The equation above is a vertical line and hence, the slope is undefined
The fourth equation is given as
![y=-2](https://img.qammunity.org/2023/formulas/mathematics/high-school/tcn3iys8rd82ithvwo5zbctm5hqvvqtks0.png)
The equation above is a horizontal line and as such,the slope is zero
Considering the graph of the equation of the line attached below
Bringing out coordinates from the graph, we will have
![\begin{gathered} (x_1,y_1)\Rightarrow(0,0) \\ (x_2,y_2)\Rightarrow(-10,5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nh3114ruioi6raz8gmplpgdd3fcvv8jpiy.png)
The slope of a line passing through points (x1,y1) and (x2,y2) is calculated using the formula below
![m=(y_2-y_1)/(x_2-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/78uaqhwt0aws3qfwxigaftpihnmb1gzxtp.png)
By substituting the values, we will have
![\begin{gathered} m=(y_2-y_1)/(x_2-x_1) \\ m=(5-0)/(-10-0) \\ m=(5)/(-10) \\ m=-(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nl1a5jr1qdzz7espcohofttuacvy6hlby5.png)
Here, the slope is = -1/2
Considering the graph of the equation of the line attached below
Bringing out coordinates from the graph, we will have
![\begin{gathered} (x_1,y_1)\Rightarrow(0,0) \\ (x_2,y_2)\Rightarrow(-2,4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r9dmnuff381e1xn6ptqybrr2qiychub26x.png)
The slope of a line passing through points (x1,y1) and (x2,y2) is calculated using the formula below
![m=(y_2-y_1)/(x_2-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/78uaqhwt0aws3qfwxigaftpihnmb1gzxtp.png)
By substituting the values, we will have
![\begin{gathered} m=(y_2-y_1)/(x_2-x_1) \\ m=(4-0)/(-2-0) \\ m=(4)/(-2) \\ m=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dleki52tdao9apcin809rq8rwsyldhsw2c.png)
Here,the slope is = -2
Therefore,
The equation with a slope of -2 is 2x +y =10
while the graph with a slope of -2 is given below