MENU
❌
Home
Math Forum/Help
Problem Solver
Practice
Algebra
Geometry
Tests
College Math
History
Games
MAIN MENU
1 Grade
Adding and subtracting up to 10
Comparing numbers up to 10
Adding and subtracting up to 20
Addition and Subtraction within 20
2 Grade
Adding and Subtracting up to 100
Addition and Subtraction within 20
3 Grade
Addition and Subtraction within 1000
Multiplication up to 5
Multiplication Table
Rounding
Dividing
Perimeter
4 Grade
Adding and Subtracting
Equivalent Fractions
Divisibility by 2, 3, 4, 5, 9
Area of Squares and Rectangles
Fractions
Equivalent Fractions
Least Common Multiple
Adding and Subtracting
Fraction Multiplication and Division
Operations
Mixed Numbers
Decimals
Expressions
6 Grade
Percents
Signed Numbers
The Coordinate Plane
Equations
Expressions
Polynomials
Symplifying Expressions
Polynomial Expressions
Factoring
7 Grade
Angles
Linear Functions
8 Grade
Linear Functions
Systems of equations
Slope
Parametric Linear Equations
Word Problems
Exponentiation
Roots
Quadratic Equations
Vieta's Formulas
Progressions
Arithmetic Progressions
Geometric Progression
Progressions
Number Sequences
Reciprocal Equations
Logarithms
Logarithmic Expressions
Logarithmic Equations
Extremal value problems
Trigonometry
Numbers Classification
Geometry
Intercept Theorem
Slope
Law of Sines
Law of Cosines
Vectors
Analytic Geometry
Probability
Limits of Functions
Properties of Triangles
Pythagorean Theorem
Inverse Trigonometric Functions
Slopes of lines: Problems with Solutions
Problem 1
What is the slope of the line?
Solution:
To get the slope of the straight line we need two points
$(x_{1},y_{1});(x_{2},y_{2})$ that the line passes through.
Points A and B have coordinates $(-4,-3)$ and $(0,3)$ and belong to the line.
The line slope formula is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
The line slope is $m=\frac{3-(-3)}{0-(-4)}=\frac{6}{4}=\frac{3}{2}$
$m=\frac{3}{2}$
Problem 2
What is the slope of the line?
$\frac{4}{5}$
$\frac{5}{4}$
$-\frac{4}{5}$
$-\frac{5}{4}$
Solution:
We need two points that a line passes through in order to find the slope of the line.
The two points are A and B with coordinates $(-2,5)$ and $(2,0)$, respectively.
The line slope formula is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
We calculate $m=\frac{0-(5)}{2-(-2)}=\frac{-5}{4}=-\frac{5}{4}$
Problem 3
What is the slope of this horizontal straight line?
Solution:
We need two points that a line passes through in order to find the slope of the line.
The two points are A and B with coordinates $(-4,4), (4,4)$ respectively.
The line slope formula is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
The line slope is $m=\frac{4-(4)}{4-(-4)}=\frac{0}{8}=0$
Problem 4
Find the slope of a straight line that passes through the points $(1,3)$ and $(-1,-1)$.
Solution:
The line slope is given by the formula $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m=\frac{-1-3}{-1-1}=\frac{-4}{-2}=2$
Problem 5
Find the slope of a line that passes through the points $(2,0)$ and $(2,3)$.
0
3
∞
1
Solution:
We will find the slope of the line using the formula
$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m=\frac{3-(0)}{2-2}=\frac{3}{0}$
The graph shows us that it is a vertical line, in this type of line its $x$ coordinate is a constant.
The equation of a vertical line is a special equation determined as $x=x_{0}$ where $x_0$ is the constant value of $x$ for all points of this line. In this special type of line we say that its slope is infinite, due to the relation between the slope and its angle of inclination with respect to the positive direction
of the $x$-axis, $m=tan(90^{\circ})=\infty$
Thus, the slope of this vertical line is infinite.
Problem 6
Find the slope of the straight line $y-1=5x+2$
Solution:
We must solve the equation for y.
$y=5x+2+1\Longrightarrow y=5x+3$
If we look at the point-slope equation of a line $y=mx+b$, we can conclude that the slope of this straight line is
$m=5$
Problem 7
Find the slope of the line $\ 2y-3x=5$
Solution:
We must solve the equation for y.
$2y=3x+5\Longrightarrow y=\frac{3}{2}x+\frac{5}{2}$.
Let we look at the point-slope equation of a line $y=mx+b$.
We can conclude in this case that the slope of this straight line is
$m=\frac{3}{2}$
Problem 8
Find the slopes $\ m_{1}$ and $m_{2}$ of the straight lines $y=3x-2$ and $x+3y+5=0$
Calculate $m_{1} \cdot m_{2} = ?$
Graph of the lines. What do you perceive?
Solution:
In the equation $x+3y+5=0$ we clear the variable $y \Longrightarrow 3y=-x-5\Longrightarrow y=-\frac{1}{3}x-\frac{5}{3}$
So we observe that $m_{1}=3,m_{2}=-\frac{1}{3}$
Now we calculate the multiplication $m_{1} \cdot m_{2}=3.(-\frac{1}{3})$
$m_{1} \cdot m_{2}=-1$
We perceive that the two lines are perpendicular.
So:
"Two straight lines are perpendicular if and only if the multiplication of their slopes is equal to $-1$"
Problem 9
Find the slopes $\ m_{1}$ and $m_{2}$ of the straight lines $y=3x-2$ and $-6x+2y+7=0$
Graph of the lines. What do you perceive?
$m_1=\frac{2}{3}, m_2=\frac{6}{7}$
$m_1=\frac{2}{3}, m_2=\frac{3}{7}$
$m_1=3, m_2=3$
$m_1=3, m_2=7$
Solution:
We must solve the equation $-6x+2y+7=0$ for y.
$2y=6x-7\Longrightarrow y=3x-\frac{7}{2}$
So we observe that $m_{1}=3,m_{2}=3$. Let's graph the lines.
They are parallel.
It is possible to demonstrate the following relationship between the parallel lines.
"Two straight lines are parallel if and only if their slopes are equal."
Problem 10
Find the equation of the line that passes through the point $(2,5)$ and its slope is $m=3$.
$y=2x+1$
$y=3x-1$
$y=3x-2$
$y=2x-3$
Solution:
We can use the point-slope form of the equation: $y-y_{1}=m(x-x_{1})$ where
$m=3$ is the slope and, $(x_{1},y_{1}) = (2, 5)$ is the point that lie to the line.
$y-5=3(x-2) \\ \Longrightarrow y=3x-6+5 \\ \Longrightarrow \fbox{y=3x-1}$
Problem 11
Find the equation of the straight line that passes through $(-2,3)$ and $(1,-2)$.
Solution:
We must find the slope of the straight line using the slope equation $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m=\frac{-2-3}{1-(-2)}=\frac{-5}{3}\Longrightarrow m=-\frac{5}{3}$
Now we will use the point-slope equation to find the equation of the line
$y-y_{0}=m(x-x_{0})$ where $(x_{0},y_{0})$ is any point that belongs to the line
$y-3=-\frac{5}{3}(x-(-2)) \Longrightarrow y-3=-\frac{5}{3}x-\frac{10}{3}\Longrightarrow $
$y=-\frac{5}{3}x-\frac{10}{3}+3=-\frac{5}{3}x-\frac{1}{3}$
Thus $y=-\frac{5}{3}x-\frac{1}{3}$
Submit a problem on this page.
Problem text:
Solution:
Answer:
Your name(if you would like to be published):
E-mail(you will be notified when the problem is published)
Notes
: use [tex][/tex] (as in the forum if you would like to use latex).
Correct:
Wrong:
Unsolved problems:
Contact email:
Follow us on
Twitter
Facebook
Author
Copyright © 2005 - 2019.