PRACTICE QUESTIONS IN SLOPE AND EQUATION OF LINES

Problem 1 :

Find the slope of the line that passes through the points (2, 4) and (6, 12)

Solution :

Using the slope formula :

Slope  =  (y₂ – y₁)/(x₂ – x₁)

(2, 4)----->(x₁, y₁)

(6, 12)----->(x₂, y₂)

  =  (12 – 4)/(6 – 2)

  =  8/4

Slope  =  2

So, the slope is 2.

Problem 2 :

Find the slope of the line.

Solution :

In the given graph, the line passes through the points (-4, 1) and (-2, -1)

Using the slope formula :

Slope  =  (y₂ – y₁)/(x₂ – x₁)

(-4, 1)----->(x₁, y₁) and (-2, -1)----->(x₂, y₂)

  =  (-1 – 1)/(-2 + 4)

  =  -2/2

Slope  =  -1

So, the slope is -1.

Problem 3 :

A climber is on a hike. After 2 hours he is at an altitude of 400 feet. After 6 hours, he is at an altitude of 700 feet. What is the average rate of changes ?

Solution :

Let, x  =  time, y  =  altitude

Then, we have the points (2, 400) and (6, 700)

We are using slope formula to find the rate of changes.

Slope formula :

Slope  =  (y₂ – y₁)/(x₂ – x₁)

(2, 400)----->(x₁, y₁)

(6, 700)----->(x₂, y₂)

  =  (700 – 400)/(6 - 2)

  =  300/4

Slope  =  75

So, the slope is 75.

Problem 4 :

At what rate did the rainfall ?

Solution :

By observing the graph, the above line is a rising line. So, it will have a positive slope.

For the above line,

Rise  =  4

Run  =  2

Then,

Slope  =  Rise/Run

=  4/2

Slope  =  2

So, the answer is 2cm per hour.

Problem 5 :

Slope of parallel lines have _____________ slope.

Solution :

So, the slope of parallel lines is the same.

Problem 6 :

Write the equation of the line

Solution :

By observing the graph, the above line is a falling line.

Slope  =  Rise/Run

=  - 3/1

Slope(m)  =  -3

Here y-intercept(b)  =  -3

Using slope-intercept form :

y  =  mx + b

y  =  -3x - 3

So, the equation of the line is y  =  -3x – 3.

Problem 7 :

y  =  3x + 2 and y  =  3x – 3

These lines are

(a) Parallel   (b) Perpendicular   (c) Neither

Solution  :

Given two lines have the same slope.

So, these lines are parallel.

Problem 8  :

Write an equation for a line parallel to the line

y  =  1/3x – 4 through (-3, 2)

Solution  :

Given, equation of the line y  =  1/3x – 4 and point (-3, 2)

Here slope(m)  =  1/3

Using point-slope form :

y – y₁  =  m(x – x₁)

(-3, 2)----->(x₁, y₁)

y – 2  =  1/3(x + 3)

y – 2  =  1/3x + 1

y  =  1/3x + 1 + 2

y  =  1/3x + 3 -----(2)

So, the equation of the line is y  =  1/3x + 3

Problem 9  :

Write an equation for a line parallel to the line

y  =  5/4x + 4 through (-4, -3)

Solution  :

Given, equation of the line y  =  5/4x + 4 and Point (-4, -3)

Here slope(m)  =  5/4

Using point-slope form :

y – y₁  =  m(x – x₁)

(-4, -3)----->(x₁, y₁)

y + 3  =  5/4(x + 4)

y + 3  =  5/4x + 5

y  =  5/4x + 5 - 3

y  =  5/4x + 2 -----(2)

So, the equation of the line is y  =  5/4x + 2.

Problem 10 :

Find the slope of the line.

Solution :

Rise  =  5 units and Run  =  3 units.

Slope  =  Rise / Run

Slope  =  5/3

Problem 11 :

Find the slope of the line.

Solution :

By observing the graph, the above line is a falling line.

Using rise over run formula :

For the above line,

Rise  =  5

Run  =  5

Slope  =  Rise/Run

Slope  =  - 5/5

Slope  =  -1

So, the answer is -1.

Problem 12 :

Find the slope of the line through each pair of points.

(-20, 13), (-15, 8)

Solution :

Given points, (-20, 13), (-15, 8)

Using the slope formula :

Slope  =  (y₂ – y₁)/(x₂ – x₁)

(-20, 13)----->(x₁, y₁)

(-15, 8)----->(x₂, y₂)

  =  (8 – 13)/(-15 + 20)

  =  -5/5

Slope  =  -1

So, the slope is -1.

Problem 13 :

Find the slope of the line through each pair of points.

(4, -1), (0, -16)

Solution :

Given points, (4, -1), (0, -16)

Using the slope formula :

Slope  =  (y₂ – y₁)/(x₂ – x₁)

(4, -1)----->(x₁, y₁)

(0, -16)----->(x₂, y₂)

  =  (-16 + 1)/(0 - 4)

  =  15/4

So, the slope is 15/4.

Problem 14  :

Write the point-slope form of the equation of the line passes through (-5, 5), parallel to y  =  -7/5x + 1

Solution :

Given point, (-5, 5) and parallel line y  =  -7/5x + 1.

Here slope(m)  =  -7/5

Using point-slope form :

y – y₁  =  m(x – x₁)

(-5, 5)----->(x₁, y₁)

y – 5  =  -7/5(x + 5)

So, the answer is y – 5  =  -7/5(x + 5).

Problem 15 :

Write the point-slope form of the equation of the line passes through (4, 1), parallel to y  =  1/2x - 4

Solution :

Given point, (4, 1) and parallel line y  =  1/2x – 4

Here slope(m)  =  1/2

Using point-slope form :

y – y₁  =  m(x – x₁)

(4, 1)----->(x₁, y₁)

y – 1  =  1/2(x + 4)

So, the answer is y – 1  =  1/2(x + 4).

Problem 16 :

Write the point-slope form of the equation of the line passes through (1, 2), perpendicular to y  =  -6/7x + 4

Solution :

Given point, (1, 2) and perpendicular line y  =  -6/7x + 4

slope(m)  =  -6/7

perpendicular slope  =  -1/m

=  -1/(-6/7)

  =  7/6

Using point-slope form :

y – y₁  =  m(x – x₁)

(1, 2)----->(x₁, y₁)

y – 2  =  7/6(x - 1)

So, the answer is y – 2  =  7/6(x - 1).

Problem 17 :

Write the point-slope form of the equation of the line passes through (-3, 5), perpendicular to y  =  1/3x + 4

Solution :

Given point, (-3, 5) and perpendicular line y  =  1/3x + 4

slope(m)  =  1/3

perpendicular slope  =  -1/m

=  -1/(1/3)

=  -3

Using point-slope form :

y – y₁  =  m(x – x₁)

(-3, 5)----->(x₁, y₁)

y – 5  =  -3(x + 3)

So, the answer is y – 5  =  -3(x + 3)

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