The general equation of a circle is given as
[tex]\begin{gathered} y=mx+c \\ \text{Where,} \\ m=\text{slope} \end{gathered}[/tex]
The first equation is given as
[tex]2x+y=10[/tex]
Making y the subject of the formula and then comparing coefficient
Subtract 2x from both sides
[tex]\begin{gathered} 2x+y=10 \\ 2x-2x+y=10-2x \\ y=-2x+10 \\ \text{slope}=-2 \end{gathered}[/tex]
The second equation is given as
[tex]-2x+y=8[/tex]
Add 2x to both sides and compare coefficients
[tex]\begin{gathered} -2x+y=8 \\ -2x+2x+y=8+2x \\ y=2x+8 \\ \text{slope}=2 \end{gathered}[/tex]
The third equation is given as
[tex]\begin{gathered} x=-2 \\ \end{gathered}[/tex]
The equation above is a vertical line and hence, the slope is undefined
The fourth equation is given as
[tex]y=-2[/tex]
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
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(0,0) \\ (x_2,y_2)\Rightarrow(-10,5) \end{gathered}[/tex]
The slope of a line passing through points (x1,y1) and (x2,y2) is calculated using the formula below
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{5-0}{-10-0} \\ m=\frac{5}{-10} \\ m=-\frac{1}{2} \end{gathered}[/tex]
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
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(0,0) \\ (x_2,y_2)\Rightarrow(-2,4) \end{gathered}[/tex]
The slope of a line passing through points (x1,y1) and (x2,y2) is calculated using the formula below
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{4-0}{-2-0} \\ m=\frac{4}{-2} \\ m=-2 \end{gathered}[/tex]
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
