Solution:
The slope of a line is given by the following equation:
![m=\text{ }(Y2-Y1)/(X2-X1)](https://img.qammunity.org/2023/formulas/mathematics/college/w955betb2uicd7nud5bfo8xxlkhoqc7mch.png)
where (X1, Y1) and (X2, Y2) are points on the line. Taking this into account, we have to:
1. Slope of the line that goes through the two points (1,3) and (7,6):
Note that in this case
(X1,Y1) = (1,3)
(X2, Y2)= (7,6)
replacing this data into the slope-equation, we get:
![m=\text{ }(Y2-Y1)/(X2-X1)\text{ = }(6-3)/(7-1)\text{ = }(3)/(6)\text{ = }(1)/(2)\text{ = 0.5}](https://img.qammunity.org/2023/formulas/mathematics/college/qtrmgn5nygbk306rmlncf6mujresx1hcf2.png)
then, the slope of the line that goes through the two points (1,3) and (7,6) is:
![m_1=\text{ }(1)/(2)\text{ = 0.5}](https://img.qammunity.org/2023/formulas/mathematics/college/w4t1v4pvchbyyz4vwojhmoo13zpoeup78e.png)
2. Slope of the line that goes through the two points (0,0) and (9,6):
Note that in this case
(X1,Y1) = (0,0)
(X2, Y2)= (9,6)
replacing this data into the slope-equation, we get:
![m=\text{ }(Y2-Y1)/(X2-X1)\text{ = }(6-0)/(9-0)\text{ = }(6)/(9)\text{ = }(2)/(3)\text{ =0.66}\approx0.7](https://img.qammunity.org/2023/formulas/mathematics/college/a47fgzjhcueojx72obsxnu4fcyp72pa5zd.png)
then, the slope of the line that goes through the two points (0,0) and (9,6) is:
![m_2=\text{ 0.66}\approx0.7](https://img.qammunity.org/2023/formulas/mathematics/college/2nhozjg2pxpin67ug7n8p42drt18cqtape.png)
note that
![m_2>m_1](https://img.qammunity.org/2023/formulas/mathematics/college/fij353r1dgys236ofnyd7uhd50wqfyd2li.png)
then, we can conclude that the second line ( the line that goes through the two points (0,0) and (9,6) ) is steeper than that the first line (the line that goes through the two points (1,3) and (7,6) ).