The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line
- Neither
- ║
- Neither
- ⊥
- ║
- Neither
- Neither
- Neither
Reason:
The slope and intercept form is the form y = m·x + c
Where;
m = The slope
Two equations are parallel if their slopes are equal
Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; [tex]m_1 = -\dfrac{1}{m_2}[/tex]
1. The given equations are in the slope and intercept form
[tex]\ y = 3 \cdot x + 1[/tex]
The slope, m₁ = 3
[tex]y = \dfrac{1}{3} \cdot x + 1[/tex]
The slope, m₂ = [tex]\dfrac{1}{3}[/tex]
Therefore, the equations are neither parallel or perpendicular
2. y = 5·x - 3
10·x - 2·y = 7
The second equation can be rewritten in the slope and intercept form as follows;
[tex]y = 5 \cdot x -\dfrac{7}{2}[/tex]
Therefore, the two equations are parallel
3. The given equations are;
-2·x - 4·y = -8
-2·x + 4·y = -8
The given equations in slope and intercept form are;
[tex]y = 2 -\dfrac{1}{2} \cdot x[/tex]
Slope, m₁ = [tex]-\dfrac{1}{2}[/tex]
[tex]y = \dfrac{1}{2} \cdot x - 2[/tex]
Slope, m₂ = [tex]\dfrac{1}{2}[/tex]
The slopes
Therefore, m₁ ≠ m₂
[tex]m_1 \neq -\dfrac{1}{m_2}[/tex]
The lines are Neither parallel nor perpendicular
4. The given equations are;
2·y - x = 2
[tex]y = \dfrac{1}{2} \cdot x +1[/tex]
m₁ = [tex]\dfrac{1}{2}[/tex]
y = -2·x + 4
m₂ = -2
Therefore;
[tex]m_1 \neq -\dfrac{1}{m_2}[/tex]
Therefore, the lines are perpendicular
5. The given equations are;
4·y = 3·x + 12
-3·x + 4·y = 2
Which gives;
First equation, [tex]y = \dfrac{3}{4} \cdot x + 3[/tex]
Second equation, [tex]y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}[/tex]
Therefore, m₁ = m₂, the lines are parallel
6. The given equations are;
8·x - 4·y = 16
Which gives; y = 2·x - 4
5·y - 10 = 3, therefore, y = [tex]\dfrac{13}{5}[/tex]
Therefore, the two equations are neither parallel nor perpendicular
7. The equations are;
2·x + 6·y = -3
Which gives [tex]y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}[/tex]
12·y = 4·x + 20
Which gives
[tex]y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}[/tex]
m₁ ≠ m₂
[tex]m_1 \neq -\dfrac{1}{m_2}[/tex]
8. 2·x - 5·y = -3
Which gives; [tex]y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}[/tex]
5·x + 27 = 6
[tex]x = -\dfrac{21}{5}[/tex]
- Therefore, the slopes are not equal, or perpendicular, the correct option is Neither
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