Point 1. To find the slope of the line you can use the following formula:
[tex]\begin{gathered} m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \text{ Where m is the slope of the line and} \\ (x_1,y_1)\text{ and}(x_2,y_2)\text{ are two points through which the line passes} \end{gathered}[/tex]If for example, you take the ordered pairs (-4,3) and (-2,4) you have
[tex]\begin{gathered} (x_1,y_1)=(-4,3) \\ (x_2,y_2)=(-2,4) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{4-3}{-2-(-4)} \\ m=\frac{1}{-2+4} \\ m=\frac{1}{2} \end{gathered}[/tex]Therefore, the slope of the line is 1/2.
Points 2 and 3. You can use the point-slope formula to find the y-intercept and to find the equation of the line in its form y = mx + b
[tex]\begin{gathered} y-y_1=m(x-x_1)\Rightarrow\text{ Point-slope formula} \\ y-3_{}=\frac{1}{2}(x-(-4)) \end{gathered}[/tex]Now, solve for y
[tex]\begin{gathered} y-3_{}=\frac{1}{2}(x+4) \\ \text{ Apply distributive property} \\ y-3_{}=\frac{1}{2}x+4\cdot\frac{1}{2} \\ y-3_{}=\frac{1}{2}x+2 \\ \text{ Add 3 from both sides of the equation} \\ y-3_{}+3=\frac{1}{2}x+2+3 \\ y=\frac{1}{2}x+5 \end{gathered}[/tex]Then, you have
[tex]\begin{gathered} y=mx+b \\ \text{ Where m is the slope of the line and} \\ \text{b is the y-intercept} \end{gathered}[/tex]Therefore, the equation of the line in its form y = mx + b is
[tex]y=\frac{1}{2}x+5[/tex]And the y-intercept is 5. Also, you can see that the line intersects the y-axis at y = 5, which confirms that the y-intercept is 5.
