Category Archives: Algebra

2.2 Lines, 2.3 Lines, Finding the Equation of a Line

Finding the Equation of a Line given two points on the line

March 18, 2017 admin

Example: Find the equation of a line in slope intercept form given the line passes through the two points (5,-3) and (6,-1) .

Solution:

First find the slope of the line.

Choose one of the points to be ( x_1, y_1) and choose the other point to be ( x_2, y_2) .

I will choose ( 5, -3) to be ( x_1, y_1) and choose ( 6, -1) to be ( x_2, y_2) .

Substitute these values into the slope formula and simplify.

$m= {y_2-y_1} / {x_2-x_1} ={-1-(-3)}/{6-5}={-1+3}/{1} =2/1=2$

The slope of the line containing the points ( 5, -3) and ( 6, -1) is m= 2 .

Then, use the point-slope formula of the line to start building the line. m represents the slope of the line and you can use (x_1,y_1) or (x_2,y_2) as the point on the line.

Point-slope formula: y-y_1 = m(x-x_1)

m=2 and (5,-3)

Substitute the values into the formula.

y-(-3) = 2(x-5)

Since the instructions ask to write the equation in slope intercept form (y=mx+b) we will simplify and write the equation with y by itself on one side.

y-(-3) = 2(x-5)

y+3 = 2x-10

y+3-3 = 5x-10-3

y = 5x-13

The equation of a line in slope intercept form passing through the two points (5,-3) and (6,-1) is y = 5x-13 .

2.2 Lines, 2.3 Lines, Finding the Equation of a Line

Finding the Equation of a Line given the slope and a point

March 18, 2017 admin

Example: Find the equation of a line in slope intercept form given the slope of the line is 7 and the line passes through the point (2,-3)

Solution:

Use the point-slope formula of the line to start building the line. m represents the slope of the line and (x_1,y_1) is a point on the line.

Point-slope formula: y-y_1 = m(x-x_1)

m=7 and (2,-3)

Substitute the values into the formula.

y-(-3) = 7(x-2)

Since the instructions ask to write the equation in slope intercept form (y=mx+b) we will simplify and write the equation with y by itself on one side.

y-(-3) = 7(x-2)

y+3 = 7x-14

y+3-3 = 7x-14-3

y = 7x-17

The equation of a line in slope intercept form with a slope of 7 and passing through the point (2,-3) is y = 7x-17 .

Example: Find the equation of the line.

2.2 Circles, 2.2 Lines, 2.3 Lines, Intercpets

X-intercepts and Y-intercepts

March 18, 2017 admin

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

In this picture, the graph crosses the x-axis at the ordered pair (2, 0) . Since every ordered pair on the x-axis has a y coordinate of zero we can let y=0 to find x-intercepts.

To find an x-intercept: Let y=0 and solve for x.

In this picture, the graph crosses the y-axis at the ordered pair (0, 6) . Since every ordered pair on the y-axis has a x coordinate of zero we can let x=0 to find y-intercepts.

To find an y-intercept: Let x=0 and solve for y.

2.2 Lines, 2.3 Lines, Graphing

Graphing Linear Equations by Finding Intercepts

March 18, 2017 admin

Steps for Graphing with the Intercept Method

Find the x intercept and the y-intercept.
- To find an x-intercept let y=0 and solve for x.
- To find a y-intercept let x=0 and solve for y.
Plot the x-intercept and y-intercept.
Draw the line that connects the intercepts.

Example: Graph the linear equation 2x-3y = 12

Solution:

1. Find the x-intercept and the y-intercept.

To find an x-intercept: Let y=0 and solve for x.

2x-3y = 12

2x-3(0) = 12

2x-0 = 12

2x = 12

{2x}/2 = {12}/2

x = 6

The x-intercept of this equation is (6,0)

To find a y-intercept: Let x=0 and solve for y.

2x-3y = 12

2(0)-3y = 12

0-3y = 12

-3y = 12

{-3y}/{-3} = {12}/{-3}

y = -4

The y-intercept of this equation is (0,-4)

2. Plot the x-intercept and the y-intercept.

3. Draw the line that connects the intercepts.

Example: Graph the linear equation y=3x

Solution:

1. Find the x-intercept and the y-intercept.

To find an x-intercept: Let y=0 and solve for x.

y=3x

0=3x

{0}/3 = {3x}/3

0 = x

The x-intercept of this equation is (0,0)

To find a y-intercept: Let x=0 and solve for y.

y=3x

y=3(0)

y=0

The y-intercept of this equation is (0,0)

Since the x-intercept and the y-intercept are the same point and we need two distinct points to graph a line, we must find another ordered pair that is a solution to the equation.

Let x=1 and find the associated y value. (I chose x=1 but you could choose a different value)

y=3x

y=3(1)

y=3

Another ordered pair on the graph is (1,3)

2. Plot the x-intercept and the y-intercept.

3. Draw the line that connects the intercepts.

Example: Graphing a linear equation with intercepts.

2.2 Lines, 2.3 Lines, Slope

Calculating Slope

March 17, 2017 admin

Given two points on the line ( x_1, y_1) and ( x_2, y_2) , you can calculate the slope of a line by the following formula.

$m= {y_2-y_1} / {x_2-x_1}$

y_2-y_1 is also know as Delta y or “the change in y.”

x_2-x_1 is also know as Delta x or “the change in x.”

$m= {y_2-y_1} / {x_2-x_1} ={Delta y} /{Delta x} ={rise}/{run}$

Example: Calculate the slope of the line containing the points ( 5, 7) and ( 9, 10) .

Solution: Choose one of the points to be ( x_1, y_1) and choose the other point to be ( x_2, y_2) .

I will choose ( 5, 7) to be ( x_1, y_1) and choose ( 9, 10) to be ( x_2, y_2) .

Substitute these values into the slope formula and simplify.

$m= {y_2-y_1} / {x_2-x_1} ={10-7}/{9-5} =3/4$

The slope of the line containing the points ( 5, 7) and ( 9, 10) is m= 3/4 .

Example: Calculate the slope of the line containing the points ( -7, -2) and ( 8, 8) .

Solution: Choose one of the points to be ( x_1, y_1) and choose the other point to be ( x_2, y_2) .

I will choose ( -7, -2) to be ( x_1, y_1) and choose ( 8, 8) to be ( x_2, y_2) .

Substitute these values into the slope formula and simplify.

$m= {y_2-y_1} / {x_2-x_1} ={8-(-2)}/{8-(-7)}={8+2}/{8+7} =10/15=2/3$

The slope of the line containing the points ( -7, -2) and ( 8, 8) is m= 2/3 .

Example: Finding the slope with the formula.

https://www.youtube.com/watch?v=1Cm7hjMUsrQ

Example: Finding the slope with the formula.

Example: Finding the slope from the graph.

2.2 Lines, 2.3 Lines, 4.1 Linear Functions and Their Properties, Slope

Interpretation of slope

March 8, 2017 admin

The slope of a line is a number that indicates the “steepness” of a line. Slope is usually denoted with the letter m.

If the slope of the line is positive, the line will be rising or increasing from left to right.

All three of the above graphs have a positive slope and the line is rising or increasing from left to right. Notice as the value of the slope gets larger, the line is getting steeper.

If the slope of the line is negative, the line will be falling or decreasing from left to right.

All three of the above graphs have a negative slope and the line is falling or decreasing from left to right. Notice as the value of the slope gets smaller, the line is getting steeper.

If the slope is zero, the line will be constant. This results in a horizontal line.

All three of the the above graphs have a slope of zero. The y values are constant.

A vertical line has a slope that is undefined.

All three of the vertical lines have undefined slope.

2.2 Circles, 2.3 Circles, Analyze

Writing an equation of a circle in standard form

March 5, 2017 admin

Example: Write the equation of a circle in standard form.

x^2+y^2+4x-8y+11=0

Start by grouping the x terms together, grouping the y terms together and moving the constant to the other side of the equation.

x^2+4x+y^2-8y=-11

Use completing the square on the group of x terms and the group of y terms.

To find the number that completes the square for the x group, start with the coefficient of the x term, half it and square it. The coefficient of the x term is 4

(4/2)^2=4

To find the number that completes the square for the y group, start with the coefficient of the y term, half it and square it. The coefficient of the y term is -8

(-8/2)^2=16

Add these numbers to the group of x terms and the group of y terms. Be careful to maintain the balance of the equation by adding the numbers to both sides of the equation.

x^2+4x+4+y^2-8y+16=-11+4+16

(x^2+4x+4)+(y^2-8y+16)=-11+4+16

Now the group of x terms is a perfect square trinomial and will factor to be a binomial squared. The group of y terms will do the same.

(x+2)(x+2)+(y-4)(y-4)=9

(x+2)^2+(y-4)^2=9

The equation of the circle is written in standard form where it is easy to recognize the center and radius of the circle.

The center is (-2,4) and the radius is 9.

2.2 Circles, 2.3 Circles, Finding Intercepts of a Circle, Intercpets

Finding the Intercepts of a Circle Touching an Axis (Tangent to an axis)

March 4, 2017 admin

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

To find an x-intercept: Let y=0 and solve for x.

To find an y-intercept: Let x=0 and solve for y.

Example: Find the intercepts of the circle for the given equation.

Solution:

To find an x-intercept, let y=0 and solve for x.

(x-1)^2+(y+2)^2=4

(x-1)^2+(0+2)^2=4

(x-1)^2+(2)^2=4

(x-1)^2+4=4

(x-1)^2+4-4=4-4

(x-1)^2=0

$sqrt{(x-1)^2}=sqrt{0}$

x-1=0

x-1+1=0+1

x=1

This equation has one x-intercept. (1,0)

To find a y-intercept, let x=0 and solve for y.

(x-1)^2+(y+2)^2=4

(0-1)^2+(y+2)^2=4

(-1)^2+(y+2)^2=4

1+(y+2)^2=4

(y+2)^2+1-1=4-1

(y+2)^2=3

$sqrt{(y+2)^2}=sqrt{3}$

y+2=pm sqrt{3}

y+2-2=-2 pm sqrt{3}

y=-2 pm sqrt{3}

Approximately y=-3.732 and x=-0.2679

This equation has two y-intercepts. (0,-2 + sqrt{3}) and (0,-2 - sqrt{3})

A tangent line to a circle may be defined as a line that intersects the circle in a single point.

This circle is tangent to the x-axis since it is touching the x-axis in a single point. The x-axis (y=0) is the tangent line for the point on the circle (1,0).

Example: Find the intercepts of the circle for the given equation.

Solution:

To find an x-intercept, let y=0 and solve for x.

(x-3)^2+(y-1)^2=9

(x-3)^2+(0-1)^2=9

(x-3)^2+(-1)^2=9

(x-3)^2+1=9

(x-3)^2+1-1=9-1

(x-3)^2=8

$sqrt{(x-3)^2}=sqrt{8}$

x-3=sqrt{4*2}

x-3=pm 2sqrt{2}

x-3+3=3 pm 2sqrt{2}

x=3 pm 2sqrt{2}

Approximately x=0.1716 and x=5.828

This equation has two x-intercepts. (3+2sqrt{2},0) and (3-2sqrt{2},0)

To find a y-intercept, let x=0 and solve for y.

(x-3)^2+(y-1)^2=9

(0-3)^2+(y-1)^2=9

(-3)^2+(y-1)^2=9

9+(y-1)^2=9

(y-1)^2+9-9=9-0

(y-1)^2=0

$sqrt{(y-1)^2}=sqrt{0}$

y-1=0

y-1+1=0+1

y=1

This equation has one y-intercept. (0,1) .

This circle is tangent to the y-axis since it is touching the y-axis in a single point. The y-axis (x=0) is the tangent line for the point on the circle (0,1).

2.2 Circles, 2.3 Circles, Finding Intercepts of a Circle, Intercpets