# SLOPE OF A LINE WORKSHEET

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.  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°.

Let (x1, y1) = (3, -2) and (x2, y2) = (-1, 4).

Then, the formula to find the slope,

m = (y2 - y1)/(x2 - x1)

Substitute (x1, y1) = (3, -2) and (x2, y2)  = (-1, 4).

m = (4 + 2)/(-1 - 3)

= - 6/4

= -3/2

So, the slope is -3/2.

Slope of the line joining (x1, y1) and (x2, y2) is,

m = (y2 - y1) / (x2 - x1)

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.

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.

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

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. 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 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 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. 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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