Category Archives: Algebra

Finding the Midpoint

The Midpoint Formula

Suppose A is ( x_1,   y_1) and B is ( x_2,    y_2)

The midpoint of the line segment joining points A and B is given by the following formula.

( {x_1+x_2}/2,   {y_1+y_2}/2)

Example: Find the midpoint of a line segment joining points A and B.

Point A is ( 4, -2) and point B is ( -8, 3)

The x coordinate of the midpoint can be found by averaging the x-values of point A and B.  The y-coordinate of the midpoint can be found by averaging the y-values of point A and B.

( {4+(-8)}/2,   {-2+3}/2)

( {-4}/2,   {1}/2)

( -2,   {1}/2)

Graphing Linear Equations with Slope Intercept Form

Example: Graph the linear equation using the slope and the y-intercept. (the slope is positive and fractional)

Example: Graph the linear equation using the slope and the y-intercept. (positive and negative fractional slopes)

Example: Graph the linear equation using the slope and the y-intercept. (equation written in standard form)

 

Finding the equation of a line perpendicular to another line

Example: Write the equation of a line in point-slope form passing through the point (-3,9) and perpendicular to the line whose equation is y={6/5}x+9/5.

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)

Although the slope of the line is not given, the slope can be deducted from the line being perpendicular to y={6/5}x+9/5.

Perpendicular lines have negative reciprocal slopes.  Since the slope of the given line is 6/5, the slope of the perpendicular line -5/6.

m=-{5/6} and  (-3,9)

Substitute the values into the point-slope formula.

y-9 = {-5/6}(x-(-3))

The point-slope form of the line is as follows.

y-9 = {-5/6}(x+3)

 

Example: Find the equation of a line perpendicular to another line and passing through a specific point. (The other line in slope intercept form)

Example: Find the equation of a line perpendicular to another line and passing through a specific point. (The other line in slope standard form)

Example: Find the equation of a line perpendicular to the x-axis.

Example: Find the equation of a line perpendicular to the x-axis and perpendicular to the y-axis.

 

Finding the Equation of a Line parallel to another line

Example: Find the equation of the line parallel to another line and passing through a specific point. (parallel equation in slope intercept form)

Example: Find the equation of the line parallel to another line and passing though a specific point. (parallel line in standard form)

Example: Find the equation of the line parallel to the x-axis or y-axis and passing through a specific point.

Example: What is an equation parallel to the y-axis?

Example: What is an equation parallel to the x-axis?

 

Finding the Equation of a line given a fractional slope and a point

Example:  Find the equation of a line in slope intercept form given the slope of the line is -{2/3} and the line passes through the point (-4,7)

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=-{2/3} and  (-4,7)

Substitute the values into the formula.

 y-7 = -{2/3}(x-(-4))

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.  I will also use the clearing fractions method to avoid having to add fractions.

 y-7 = -{2/3}(x-(-4))

3(y-7) = 3[-{2/3}(x+4)] (Multiply by LCM)

3(y-7) = -2(x+4) (Cancel Denominator)

3y-21 = -2x-8

3y-21+21 = -2x-8+21

3y = -2x+13

{3y}/3 = {-2x}/3+{13}/3

y = -{2/3}x+{13}/3

The equation of a line in slope intercept form with a slope of -{2/3} and  passing through the point (-4,7) is   y = -{2/3}x+{13}/3