SOLUTION
Given the information in the image on the question tab;
We have to form triangles using points A and B, and points A and C;
Then find the slopes between A and B, and A and C;
Part A;
[tex]\begin{gathered} Slope\text{ between point A and B is:} \\ \frac{AM}{BM}=\frac{147-105}{2-1}=\frac{42}{1}=42 \end{gathered}[/tex][tex]\begin{gathered} Slope\text{ between point A and C is:} \\ \frac{AN}{CN}=\frac{147-63}{2-0}=\frac{84}{2}=42 \end{gathered}[/tex]
The slopes are equal.
Part B;
[tex]The\text{ initial value of the graph is 63m}[/tex]
The initial value, or y-intercept, is the output value when the input of a linear function is zero. It is the y-value of the point at which the line crosses the y-axis.
[tex]The\text{ slope of the graph is 42.}[/tex]
The slope represents change in y over change in x. Therefore, in a distance vs. time graph, the slope is equal to distance/time, which represents the velocity.
