1. Writing equations given slope and point
5.2
12/6/13

2. Solving Equations
In any equation, we can solve for any variable that is not known
Rewrite in terms of that variable
Find a value if all the other values are known
Can use this to find equations

3. Point-Slope Form of the Equation of a Line
The point-slope equation of a non-vertical line of slope m that passes through the point (x1, y1) is
y – y1 = m(x – x1).

4. Solving With Slope and a Point
Substitute known values into the equation
Rewrite the equation y = mx + b with slope and y-intercept values

5. Example: Writing the Point-Slope Equation of a Line
Write the point-slope form of the equation of the line passing through (-1,3) with a slope of 4. Then solve the equation for y.
Solution We use the point-slope equation of a line with m = 4, x1= -1, and y1 = 3.
This is the point-slope form of the equation.
y – y1 = m(x – x1)
Substitute the given values. Simply.
y – 3 = 4[x – (-1)]
We now have the point-slope form of the equation for the given line.
y – 3 = 4(x + 1)
We can solve the equation for y by applying the distributive property.
y – 3 = 4x + 4
y = 4x + 7
Add 3 to both sides.

6. Solving with Slope and Point
Find the equation of the line passing through the point (-3, 0 ) and has slope of 1/3

7. Solving with Slope and a Point
Find the equation of the line passing through point (-2, -1) with slope of -3

8. Solving with Slope and Point
Find the equation of the line passing through the point (-3, 0 ) and has slope of 1/3

9. Solving with Slope and a Point
Find the equation of the line passing through (3 , -4) and is parallel to the line y = -3x – 2

10. Slope and Perpendicular Lines
90°
Two lines that intersect at a right angle (90°) are said to be perpendicular. There is a relationship between the slopes of perpendicular lines.
Slope and Perpendicular Lines
If two non-vertical lines are perpendicular, then the product of their slopes is –1.
If the product of the slopes of two lines is –1, then the lines are perpendicular.
A horizontal line having zero slope is perpendicular to a vertical line having undefined slope.

11. Perpendicular Lines
Perpendicular lines have negative reciprocal slopes so if you need the slope of a line perpendicular to a given line, simply find the slope of the given line, take its reciprocal (flip it over) and make it negative.

12. Example: Finding the Slope of a Line Perpendicular to a Given Line
Find the slope of any line that is perpendicular to the line whose equation is x + 4y – 8 = 0.
Solution We begin by writing the equation of the given line in slope-intercept form. Solve for y.
x + 4y – 8 = 0
This is the given equation.
4y = -x + 8
To isolate the y-term, subtract x and add 8 on both sides.
Slope is –1/4.
y = -1/4x + 2
Divide both sides by 4.
The given line has slope –1/4. Any line perpendicular to this line has a slope that is the negative reciprocal, 4.

13. Example: Writing the Equation of a Line Perpendicular to a Given Line
Write the equation of the line perpendicular to x + 4y – 8 = 0 that passes thru the point (2,8) in standard form.
Solution: The given line has slope –1/4. Any line perpendicular to this line has a slope that is the negative reciprocal, 4.
So now we need know the perpendicular slope and are given a point (2,8). Plug this into the point-slope form and rearrange into the standard form.
y – 8 = 4[x – (2)]
y – y1 = m(x – x1)
y1 = 8
m = 4
x1 = 2
y - 8 = 4x - 8
-4x + y = 0
4x – y = 0 Standard form

14. 4 -1 2 y = - x 1 Let's look at a line and a point not on the line
Let's find the equation of a line parallel to y = - x that passes through the point (2, 4)
y = - x
What is the slope of the first line, y = - x ?
(2, 4) 1
This is in slope intercept form so y = mx + b which means the slope is –1.
So we know the slope is –1 and it passes through (2, 4). Having the point and the slope, we can use the point-slope formula to find the equation of the line 4 -1 2
Distribute and then solve for y to leave in slope-intercept form.

15. 4 1 2 y = - x What if we wanted perpendicular instead of parallel?
Let's find the equation of a line perpendicular to y = - x that passes through the point (2, 4)
y = - x
The slope of the first line is still –1.
(2, 4)
The slope of a line perpendicular is the negative reciporical so take –1 and "flip" it over and make it negative. 4 1 2
Distribute and then solve for y to leave in slope-intercept form.
So the slope of a perpendicular line is 1 and it passes through (2, 4).

16. Homework:
5.2 worksheet #’s 1-15
Due Monday, 12/9