1. 6.5 Slope-Point Form of the Equation for a Linear Function
Linear Functions
6.5 Slope-Point Form of the Equation for a Linear Function

2. Today’s Objectives
Relate linear relations expressed in: slope-intercept form, general form, and slope-point form to their graphs, including:
Express a linear relation in different forms, and compare graphs
Rewrite a linear relation in either slope-intercept or general form
Graph, with or without technology, a linear relation in slope-intercept, general, or slope-point form
Identify equivalent linear relations from a set of linear relations
Match a set of linear relations to their graphs

3. Equations of a Linear Function
We can make an equation that describes a line’s location on a graph. This is called a linear equation. There are three forms of linear equation that we will be looking at:
Standard Form: Ax + By + C = 0, where A, B, and C are integers.
Slope y-intercept form: y = mx + b, where m is the slope, and b is the y-intercept.
Slope-point form: y – y1 = m(x – x1), where m is the slope, and the line passes through a point located at (x1, y1)

4. Slope-point form
When we know the slope of a line and the coordinates of a point on the line, we use the property that the slope of a line is constant to determine an equation for the line.
In this example, the line has a slope of -3 and passes through . We use any other point on the line to write an equation for the slope, m:
(-2,5)

5. Example
A line has a slope of -3 and passes through P(-2,5). Write an equation in slope-point form. Slope = rise/run m = y - 5 / x - (-2) -3 = y - 5 / x (x+2) = y – 5 So y – 5 = -3(x+2) {slope-point form}

6. Example:
Describe the graph of the equation in slope-point form, then graph the equation
𝐲−𝟐= 𝟏 𝟑 𝐱+𝟒
Solution:
From the equation, we can see the x-value of the point P, is at -4, and the y coordinate of the point P is at 2. So, point P is found at (-4, 2). Also, we can see that the slope is 1/3. We can use this information to now graph the line.

7. Example Writing an Equation Using a Point on the Line and its Slope
A) Write an equation in slope-point form for this line:
B) Write the equation in part A in slope-intercept form. What is the y-intercept of this line?

8. Example 2 - Solution
A) Count out the rise and run to determine the slope of the line.
Use the slope-point form of the equation:
Substitute a point easily read from the graph and our slope: (-1,-2):
B) First, we can remove the brackets by multiplying:
Next, we solve for y:
As we can see from this equation, the y-intercept is at:

9. Parallel and Perpendicular Lines
Writing the equations of parallel and perpendicular lines in slope-point form is very similar to doing the same in slope-intercept form.
The only difference between the equations is in the value of m, the slope.
Recall that the slope of a line perpendicular to another line is the negative reciprocal of the original line.

10. Example
Write an equation for the line that passes through R(1,-1) and is:
A) Parallel to the line
B) Perpendicular to the line
Solution:
First, we can see from the lines that the slope is 2/3. Any line parallel to this line has slope 2/3. Now, we use our point R(1,-1) to find the equation of the line:
Any line perpendicular to this line has a slope that is the negative reciprocal of 2/3, which is -3/2. So, the equation of a line perpendicular to the line is:

11. Example:
Write an equation for the line that passes through S(2,-3) and is:
A) Parallel to the line y = 3x + 5
B) Perpendicular to the line y = 2/3x - 5
Solution:
First, we can see from the lines that the slope is 3. Any line parallel to this line has slope 3.
Now, we use our point to find the equation of the line:
y – (-3) = 3(x – 2)
y + 3 = 3(x-2)
Any line perpendicular to this line has a slope that is the negative reciprocal of 3, which is -1/3. So, the equation of a line perpendicular to the line is:
y + 3 = -1/3(x-2)

12. Using 2 Points to Write an Equation
We can use the coordinates of two points that satisfy a linear function P(x1,y1) and Q(x2,y2) to write an equation for the function.
We write the slope of the graph of the function in two ways:
So, an equation is:

13. Example Given K(1,-2) and J(6,8)
Determine the slope-point form of the equation.
Calculate the slope of the line.
m = 8-(-2)/ m = 2
Use either point K or J to write the slope-intercept form of the equation.
Using K(1,-2) Using J(6,8)
y – (-2) = 2(x -1)
y + 2 = 2(x -1) y – 8 = 2(x -6)
y = 2x - 2 – y = 2x –
y = 2x – y = 2x – 4

14. Example Write the equation of a Linear Function Given Two Points
The sum of angles, s degrees, in a polygon is a linear function of the number of sides, n, of the polygon. The sum of the angles in a triangle (3 sides) is 180º. The sum of the angles in a quadrilateral (4 sides) is 360º.
A) Write the linear equation to represent this function
B) Use the equation to determine the sum of angles in a dodecagon (shape with 12 sides)
HINT: Determine your x and y axis ? ? ? ? ?
number of side [n] is the independent variable (x)
sum of angles [s] is the dependent variable (y)
Determine the coordinates of the two points given.
Determine the slope.

15. Example 4 Solution: A) Slope-point form: Slope y-intercept form: B)
The sum of the angles in a dodecagon is 1800º