graphing linear equations, point- slope form, and parallel/perpendicular lines review algebra honors...

Graphing Linear Equations, Point-Slope Form, and

Parallel/Perpendicular linesREVIEW

Algebra HonorsMr Smith

Objective

• By the end of this lesson you should be able to take given points or slope and graph and write it in point-slope, slope-intercept, and standard forms.

• By the end of this lesson you should be able to write equations for parallel and perpendicular lines.

Slope- Intercept form

• To _______, it is best to have the linear equation in slope-intercept form.

• Slope intercept form is y = ___ + ___• The 4 variables and definitions of slope int

form are:___ = _______ ____ = ___________ = _______ ____ = ________Pause!

Slope- Intercept form

• To _Graph_, it is best to have the linear equation in slope-intercept form.

• Slope intercept form is y = mx + b• The 4 variables and definitions of slope int

form are:_y_ = total _m_ = slope_ x = units_ b = y - intercept

Writing in slope- intercept

• Slope = 5, y-int = -7

• 4x + 3y = 18 (-4, 6) (3, -8)

Pause!

Writing and Graphing in slope- intercept form

1. Slope = 5, y-int = -7 y = 5x - 72. 4x + 3y = 18 3. (-4, 6) (3, -8)= -4x -4x m = = -2

3y = -4x + 18 y=mx+b 3 3 6=-2(-4)+b

y = 6=8+bb = -2 y=-2x-2

Graphing

For the following, the first slide is the problem, the second is the solution

The first step, which we just did, is to put the info into slope-int form, and then graph.

Now, Graph y = 5x - 7

Graph y = 5x - 7

Graph y =

Graph y =

Graph y=-2x-2

Now, Graph y=-2x-2

Point-Slope Form

• You can write the equation of a line using point slope form, even if you do not know the second point on a line.

Write in point-slopey – y1 = m (x – x1)

A line passing through point (4, -5) with a slope of -2

Pause!

Write in point-slopey – y1 = m (x – x1)

A line passing through point (4, -5) with a slope of -2y + 5 = -2 (x – 4)

Next, write this in slope- int, and then standard form

Pause!

To slope – int To standard

y + 5 = -2 (x – 4) y = -2x + 3y + 5 = -2x + 8 +2x +2x -5 -5 2x + y = 3 y = -2x + 3

Write in point-slopey – y1 = m (x – x1)

A line passing through (3, -4) and (-6, -1)

Pause!

Write in point-slopey – y1 = m (x – x1)

A line passing through (3, -4) and (-6, -1)1. Find the slope = =

next, put into slope-int and standard

2. Pick a point and the slope y - -4 = (x – 3) y + 4 = (x – 3)

Pause!

To slope – int To standard

y + 4 = (x – 3) y = x + 3 + x +

y + 4 = x + 1 (3) x + y = 3(3) -4 -4 y = x + 3 x + 3y = 9

Parallel and Perpendicular

Parallel lines Have the same slope, but different y-intercepts

Perpendicular Lines have Opposite Reciprocal slopes, but could have the same intercept.

Example: a slope of has opposite reciprocal slope of -

Write the equation of a line that is…

Parallel to y = 5x + 6, and goes through (-5, 9)Slope is the same, so m=5, solve for b:y = mx+b > 9 = 5(-5) + b 9 = -25 + b

+25 +25 34 = b, so …

y = 5x + 34 is the answer

Write the equation of a line that is…

Perpendicular to y=5x+6, and goes through(-3,6)Slope is opposite reciprocal, so m = - y = mx + b > 6 = - (-3) + b > 6 = + b

- - 5 = b, so… y = - + 5

Yes, you can have intercepts that are not integers

Things to Remember

• To Graph, you should use slope-intercept form. Start at the y-intercept, and then plot the slope from there.

• Moving between forms really comes down to moving terms around, and watching your signs.

• Parallel lines have the same slope, perpendicular lines are opposite reciprocals.