How to decide whether the lines are parallel, perpendicular or neither ?
Parallel lines :
Parallel lines will have same slope and different y-intercepts.
m_{1} = m_{2}
m_{1} and m_{2} are the slope of those two lines.
Perpendicular lines :
If two lines are perpendicular then product of their slopes will be equal to -1.
m_{1}.m_{2 }= -1
Coincident lines :
Coincident lines will have same slope and same y-intercepts.
The lines which has no relationship in between the slopes and y-intercepts can be considered neither.
Tell whether the lines are parallel, perpendicular, or neither.
Problem 1 :
Line 1 : through (-3, -7) and (1, 9)
Line 2 : through (-1, -4) and (0, -2)
Solution :
Slope of line 1 : x_{1} = -3, y_{1} = -7 x_{2} = 1, y_{2} = 9 m_{1} = (y_{2} – y_{1})/(x_{2} – x_{1}) = (9 + 7)/(1 + 3) = 16/4 = 4 |
Slope of line 2 : x_{1} = -1, y_{1} = -4 x_{2} = 0, y_{2} = -2 m = (y_{2} – y_{1})/(x_{2} – x_{1}) = (-2 + 4)/(0 + 1) = 2/1 = 2 |
They are not parallel, they are not perpendicular. So, the lines are neither.
Problem 2 :
Line 1 : through (2, 7) and (-1, -2)
Line 2 : through (3, -6) and (-6, -3)
Solution :
Slope of line 1 : x_{1} = 2, y_{1} = 7 x_{2} = -1, y_{2} = -2 m_{1} = (y_{2} – y_{1})/(x_{2} – x_{1}) = (-2 - 7)/(-1 - 2) = -9/(-3) = 3 |
Slope of line 2 : x_{1} = 3, y_{1} = -6 x_{2} = -6, y_{2} = -3 m_{2} = (y_{2} – y_{1})/(x_{2} – x_{1}) = (-3 + 6)/(-6 - 3) = 3/-9 = -1/3 |
m_{1} x m_{2} = 3 (-1/3)
m_{1} x m_{2} = -1
So, the lines are perpendicular.
Determine whether
the graphs of each pair of equations are parallel, perpendicular, or neither.
Problem 3 :
y = 3x + 4
y = 3x + 7
Solution :
y = 3x + 4 ----- (1)
y = 3x + 7 ------(2)
Comparing the above equations with slope intercept form
y = mx + b, we get
m_{1} = 3, m_{2} = 3
b_{1} = 4, b_{2} = 7
Both lines are having same slope. but different y – intercepts. So, the given lines are parallel.
Problem 4 :
y = -4x + 1
4y = x + 3
Solution :
y = -4x + 1 ------(1)
4y = x + 3
y = x/4 + 3/4 ------(2)
m = -4, m = 1/4
m_{1} ⋅ m_{2} = -4 (1/4)
m_{1} ⋅ m_{2} = -1
Since the product of the slopes is equal to -1. The lines are perpendicular.
Problem 5 :
y = 2x - 5
y = 5x - 5
Solution :
y = 2x – 5 -----(1)
y = 5x - 5 -----(2)
Comparing (1) and (2)
y = mx + b
m = 2, m = 5
b = -5, b = -5
There is no relationship between slopes, so they are neither.
Problem 6 :
y = -1/3x + 2
y = 3x - 5
Solution :
y = -1/3x + 2 -----(1)
y = 3x - 5 -----(2)
y = mx + b
m_{1} = -1/3, m_{2} = 3
b = 2, b = -5
m_{1} = -1/3, m_{2} = 3
m_{1 }⋅ m_{2 }= (-1/3) ⋅ 3
Product of the slope is equal to -1. So, the lines are perpendicular.
Problem 7 :
y = 3/5x - 3
5y = 3x - 10
Solution :
y = 3/5x – 3 ------(1)
5y = 3x - 10
y = 3/5x - 10 ------(2)
y = mx + b
m_{1} = 3/5, m_{2} = 3/5
b = -3, b = -10
Both lines have the same slope but have different y-intercepts. So, the lines are parallel.
Problem 8 :
y = 4
4y = 6
Solution :
y = 4 ------(1)
4y = 6
y = 6/4
y = 3/2------(2)
Both are horizontal lines and they will not intersect.
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