Answer:
Problem 10:
Slope of line KI [tex]=\boxed{2}\frac{\boxed{2}}{\boxed{3}}[/tex]
Problem 11:
Slope of line BC [tex]=\boxed{-}\frac{\boxed{2}}{\boxed{5}}[/tex]
Step by step explanation:
Problem 10:
- Line is passing through the points I(-3, 5) & K(-5, -3)
- [tex]\implies x_1 = -3,\: y_1= 5,\: x_2=-5, \: \&\: y_2 =-3[/tex]
- Slope of line KI [tex]=\frac{y_2-y_1}{x_2-x_1}[/tex]
- Slope of line KI [tex]=\frac{-3-5}{-5-(-3)}[/tex]
- Slope of line KI [tex]=-\frac{-8}{-5+2}[/tex]
- Slope of line KI [tex]=\frac{-8}{-3}[/tex]
- Slope of line KI [tex]=\boxed{2}\frac{\boxed{2}}{\boxed{3}}[/tex]
Problem 11:
- Line is passing through the points B(-1, 2) & C(4, 0)
- [tex]\implies x_1 = -1,\: y_1= 2,\: x_2=4, \: \&\: y_2 =0[/tex]
- Slope of line BC [tex]=\frac{y_2-y_1}{x_2-x_1}[/tex]
- Slope of line BC [tex]=\frac{0-2}{4-(-1)}[/tex]
- Slope of line BC [tex]=-\frac{2}{4+1}[/tex]
- Slope of line BC [tex]=-\frac{2}{5}[/tex]
- Slope of line BC [tex]=\boxed{-}\frac{\boxed{2}}{\boxed{5}}[/tex]
