Find the slope of the following straight lines :
Example 1 :
5y - 3 = 0
Solution :
5y - 3 = 0
5y = 3
y = 3/5
This is the equation of an horizontal line passing through the value 3/5 on the y-axis.
We know that the slope of an horizontal line is always zero.
Hence the slope of the given line is 0.
Example 2 :
7x - 3/17 = 0
Solution :
7x - 3/17 = 0
7x = 3/17
x = 3/119
This is the equation of a vertical line passing through the value 3/119 on the x-axis.
We know that the slope of a vertical line is always undefined.
Hence the slope of the given line is undefined.
Example 3 :
Find the slope of the line which is
(i) parallel to y = 0.7x - 11
(ii) perpendicular to the line x = -11
Solution :
Part (i) :
y = 0.7x - 11
The given equation of the line is in slope-intercept form.
Comparing y = mx + b and y = 0.7x - 11, we get
m = 0.7
Slope of the given line = 0.7
If the lines are parallel, then the slopes must be equal.
Then, the slope of the line parallel to y = 0.7x - 11 is also 0.7.
Part (ii) :
x = -11
This is the equation of a vertical line passing through the value -11 on the x-axis.
We know that the slope of a vertical line is always undefined.
So, the slope of the line x + 11 = 0 is undefined.
Let k be the slope of the line perpendicular to x + 11 = 0.
If the lines are perpendicular, then the product of the slopes must be equal to -1.
k x undefined = -1
= -1/undefined
= 0
Then, the slope of the line perpendicular to x = -11 is zero.
Example 4 :
Check whether the following two lines are parallel.
x/3 + y/4 + 1/7 = 0
2x/3 + y/2 + 1/10 = 0
Solution :
x/3 + y/4 + 1/7 = 0 ----(1)
2x/3 + y/2 + 1/10 = 0 ----(2)
Let m_{1} and m_{2} be the slopes of the lines (1) and (2) respectively.
Find the slope of line (1) :
x/3 + y/4 + 1/7 = 0
y/4 = -x/3 - 1/7
y = (-4/3)x - 4/7
Comparing y = mx + b and y = (-4/3)x - 4/7, we get
m = -4/3
Slope of line (1) = -4/3
m_{1} = -4/3
Find the slope of line (2) :
2x/3 + y/2 + 1/10 = 0
y/2 = -2x/3 - 1/10
y = (-4/3)x - 2/10
y = (-4/3)x - 1/5
Comparing y = mx + b and y = (-4/3)x - 1/5, we get
m = -4/3
Slope of line (2) = -4/3
m_{2} = -4/3
m_{1} = m_{2}
Since the slopes are equal, the lines are parallel.
Example 5 :
Check whether the following two lines are perpendicular.
5x + 23y + 14 = 0
23x − 5y + 9 = 0
Solution :
Let m_{1} and m_{2} be the slopes of the lines (1) and (2) respectively.
Find the slope of line (1) :
5x + 23y + 14 = 0
23y = -5x - 14
y = (-5/23)x - 14/23
Comparing y = mx + b and y = (-5/23)x - 14/23, we get
m = -5/23
Slope of line (1) = -5/23
m_{1} = -5/23
Find the slope of line (2) :
23x - 5y + 9 = 0
-5y = -23x - 9
5y = 23x + 9
y = (23/5) + 9/5
Comparing y = mx + b and y = (23/5)x + 9/5, we get
m = 23/5
Slope of line (2) = 23/5
m_{2} = 23/5
m_{1} ⋅ m_{2 }= (-5/23)(23/5)
m_{1} ⋅ m_{2 }= -1
Since the product of the slopes is equal to -1, the lines are perpendicular.
Example 6 :
If the following two lines are perpendicular, then find the value of p.
12y = -(p + 3)x + 12
12x - 7y = 16
Solution :
12y = -(p + 3)x + 12 ----(1)
12x - 7y = 16 ----(2)
Find the slope of line (1) :
12y = -(p + 3)x + 12
y = [-(p + 3)/12]x + 12/12
y = [-(p + 3)/12]x + 1
Comparing y = mx + b and y = [-(p + 3)/12]x + 1, we get
m = -(p + 3)/12
Slope of line (1) = -(p + 3)/23
Find the slope of line (2) :
12x - 7y = 16
-7y = -12x + 16
7y = 12x - 16
y = (12/7)x - 16/7
Comparing y = mx + b and y = (12/7)x - 16/7, we get
m = 12/7
Slope of line (2) = 12/7
Since the lines are perpendicular, product of the slopes is equal to -1.
[-(p + 3)/12][12/7] = -1
-(p + 3)/7 = -1
p + 3 = 7
p = 4
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Oct 05, 22 11:41 AM
Oct 04, 22 11:08 PM
Oct 04, 22 11:07 PM