MATH 1201 College Algebra.
MATH 1201 College Algebra.
Written Assignment Unit 2.
1. Determine whether the lines given by the equations below are parallel, perpendicular, or
neither. Also, find a rigorous algebraic solution for each problem.
a. {3y+4x=12 \brace -6y=8x+1}
Solution.
3y = 12 – 4x
-6y = 8x + 1
Making y the subject in both cases:
y = 12/3 – 4/3x
y = 4 – 4/3x
y = -4/3x + 4 …………………. (1)
y = -8/6x -1/6
y = -4/3x – 1/6 ……………. (2)
Using the general formula for a linear function y = mx + c, the two functions have the same
gradient, m = -4/3. Since the gradients for the two lines are the same, we conclude that the two
lines are parallel to each other.
The graphs of the two lines are shown:
1|Brian Wanyonyi- University of the People.
, MATH 1201 College Algebra.
b. {3y+x=12 \brace -y=8x+1}
3y = 12 – x
y= 4 -1/3x
y = -1/3x + 4
m1 = -1/3 ……………… (1)
For the second equation:
-y= 8x + 1
y= -8x -1
m2= -8 …………………. (2)
The two equations are neither parallel nor perpendicular. The gradients are not the same, hence
not parallel. The product of the gradients is not equal to -1, hence no perpendicular lines.
The graph for these two lines is as shown below:
c. {4x-7y=10 \brace 7x+4y=1}
-7y = 10 – 4x
y = -10/7 + 4/7X
2|Brian Wanyonyi- University of the People.
MATH 1201 College Algebra.
Written Assignment Unit 2.
1. Determine whether the lines given by the equations below are parallel, perpendicular, or
neither. Also, find a rigorous algebraic solution for each problem.
a. {3y+4x=12 \brace -6y=8x+1}
Solution.
3y = 12 – 4x
-6y = 8x + 1
Making y the subject in both cases:
y = 12/3 – 4/3x
y = 4 – 4/3x
y = -4/3x + 4 …………………. (1)
y = -8/6x -1/6
y = -4/3x – 1/6 ……………. (2)
Using the general formula for a linear function y = mx + c, the two functions have the same
gradient, m = -4/3. Since the gradients for the two lines are the same, we conclude that the two
lines are parallel to each other.
The graphs of the two lines are shown:
1|Brian Wanyonyi- University of the People.
, MATH 1201 College Algebra.
b. {3y+x=12 \brace -y=8x+1}
3y = 12 – x
y= 4 -1/3x
y = -1/3x + 4
m1 = -1/3 ……………… (1)
For the second equation:
-y= 8x + 1
y= -8x -1
m2= -8 …………………. (2)
The two equations are neither parallel nor perpendicular. The gradients are not the same, hence
not parallel. The product of the gradients is not equal to -1, hence no perpendicular lines.
The graph for these two lines is as shown below:
c. {4x-7y=10 \brace 7x+4y=1}
-7y = 10 – 4x
y = -10/7 + 4/7X
2|Brian Wanyonyi- University of the People.
