Problem 1 :
Find the angle of inclination of the straight line whose slope is 1/√3.
Problem 2 :
Find the slope of the straight line passing through the points (3, -2) and (-1, 4).
Problem 3 :
Using the concept of slope, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
Problem 4 :
Find the slope of the line 3x - 2y + 7 = 0.
Problem 5 :
If the straight line 5x + ky - 1 = 0 has the slope 5, find the value of 'k'.
Problem 6 :
If the straight kx - 2y + 3 = 0 is passing through the (1, 3), find its slope.
Problem 7 :
Find the slope of the line shown below using rise over run formula.
Problem 8 :
Find the slope of the line shown below using rise over run formula.
Problem 9 :
Find the slope of the line shown below using rise over run formula.
Problem 10 :
Find the slope of the line shown below using rise over run formula.
1. Answer :
Let θ be the angle of inclination of the line.
Then, slope of the line is
m = tan θ
Given : Slope = 1/√3.
tan θ = 1/√3
θ = 30°
So, the angle of inclination is 30°.
2. Answer :
Let (x_{1}, y_{1}) = (3, -2) and (x_{2}, y_{2}) = (-1, 4).
Then, the formula to find the slope,
m = (y_{2} - y_{1})/(x_{2} - x_{1})
Substitute (x_{1}, y_{1}) = (3, -2) and (x_{2}, y_{2}) = (-1, 4).
m = (4 + 2)/(-1 - 3)
= - 6/4
= -3/2
So, the slope is -3/2.
3. Answer :
Slope of the line joining (x_{1}, y_{1}) and (x_{2}, y_{2}) is,
m = (y_{2} - y_{1}) / (x_{2} - x_{1})
Using the above formula, slope of the line AB joining the points A(5, -2) and B(4, -1) is
= (-1 + 2)/(4 - 5)
= -1
Slope of the line BC joining the points B(4, -1) and C(1, 2) is
= (2 + 1)/(1 - 4)
= -1
Thus,
slope of AB = slope of BC
Also, B is the common point.
So, the points A, B and C are collinear.
4. Answer :
Equation of the given line :
3x - 2y + 7 = 0
Write the above equation in slope intercept form, that is
y = mx + b
where m is the slope and b is the intercept.
3x - 2y + 7 = 0
-2y = -3x - 7
2y = 3x + 7
y = 3x/2 + 7/2
y = (3/2)x + 7/2
The above equation is in slope intercept form.
m = 3/2
So, the slope of the given line is 3/2.
5. Answer :
Equation of the given line :
5x + ky - 1 = 0
Write the above equation in slope intercept form.
5x + ky - 1 = 0
ky = -5x + 1
y = -5x/k + 1/k
y = (-5/k)x + 1/k
The above equation is in slope intercept form.
m = -5/k
Given : Slope = 5.
5 = -5/k
5k = -5
k = -1
6. Answer :
Because the straight line kx - 2y + 3 = 0 is passing through the point (1, 3), we can substitute x = 1 and y = 3 into the equation.
k(1) - 2(3) + 3 = 0
k - 6 + 3 = 0
k - 3 = 0
k = 3
Equation of the given straight line :
x - 2y + 3 = 0
Write the above equation in slope intercept form.
x - 2y + 3 = 0
-2y = -x - 3
2y = x + 3
y = x/2 + 3/2
y = (1/2)x + 3/2
The above equation is in slope intercept form.
m = 1/2
So, the slope of the given line is 1/2.
7. Answer :
The above line is a rising line. So, its slope will be a positive value.
Measure the rise and run.
For the above line,
rise = 2
run = 5
Then,
slope = rise/run
slope = 2/5
8. Answer :
The above line is a falling line. So, its slope will be a negative value.
Measure the rise and run.
For the above line,
rise = 7
run = 9
Then,
slope = rise/run
slope = -7/9
9. Answer :
The above line is a vertical line.
Measure the rise and run.
For the above line,
rise = 3
run = 0
Then,
slope = rise/run
slope = 3/0
slope = undefined
Note :
The slope of a vertical line is always undefined.
10. Answer :
The above line is an horizontal line.
Measure the rise and run.
For the above line,
rise = 0
run = 4
Then,
slope = rise/run
slope = 0/4
slope = 0
Note :
The slope of an horizontal line is always zero.
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