Answer:
Distance = 5 units
Slope = [tex]\displaystyle\frac{4}{3}[/tex]
Step-by-step explanation:
Hi there!
1) Finding the distance of the line
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex] where two endpoints of a line are [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]
Plug in the points (-7,2) and (-4,6):
[tex]d=\sqrt{(-4-(-7))^2+(6-2)^2}\\d=\sqrt{(-4+7)^2+(6-2)^2}\\d=\sqrt{(3)^2+(4)^2}\\d=\sqrt{9+16}\\d=\sqrt{25}\\d=5[/tex]
Therefore, the length of the line/distance of the line is 5 units.
2) Finding the slope of the line
[tex]m=\displaystyle\frac{y_2-y_1}{x_2-x_1}[/tex] where two points are [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]
Plug in the points (-7,2) and (-4,6):
[tex]m=\displaystyle\frac{6-2}{-4-(-7)}\\\\m=\displaystyle\frac{6-2}{-4+7}\\\\m=\displaystyle\frac{4}{3}[/tex]
Therefore, the slope of the line is [tex]\displaystyle\frac{4}{3}[/tex].
I hope this helps!
