Writing Equations of Lines Chapter 2 Section 4 Writing Equations of Lines
Writing An Equation of a Line Slope-Intercept Form: Given the slope m and the y-intercept b, use this equation: f(x) = m x + b Point Slope Form: Given the slope m and a point (x1, y1), or given two points, (x1, y1), and (x2, y2), use this equation: f(x) – y1 = m (x – x1)
Point – Slope Form To write an equation of a line in point – slope form, all you need is … … Any Point On The Line … (x1, y1) … The Slope … m Once you have these two things, you can write the equation as f(x) – y1 = m (x – x1) That’s “y minus the y-value of the point equals the slope times the quantity of x minus the x-value of the point”.
From the graph you can see that Example x y +2 +3 Write an equation of the line shown. From the graph you can see that m = b = -3 Use f(x) = mx + b So the equation is
Example Write the equation of the line that goes through the point (2, 3) and has a slope of -1/2. Point = (2, 3) Slope = -1/2 Starting with the point – slope form f(x) – y1 = m (x – x1) Plug in the y-value, the slope, and the x-value to get f(x) - 3 = -1/2 (x – 2) f(x) – 3 = -1/2x + 1 f(x) = -1/2x + 4
Graphing Graph your result:
Parallel Lines have slopes that are the same. Perpendicular Lines have slopes that are opposite reciprocals.
Example Write an equation of the line that passes through (3, 2) and is parallel to f(x) = -3x +2
Example Write an equation of the line that passes through (3, 2) and is perpendicular to f(x) = -3x +2
Graph the results Original Line f(x) = -3x + 2 Parallel Line Perpendicular Line f(x) = 1/3 x + 1
Using the first point, we have, Example Write the equation of the line that goes through the points (6, –4) and (2, 8) . We have two points, but we’re missing the slope. Using the formula for slope, we can find the slope to be f(x)2 – f(x)1 x2 – x1 To use point – slope form, we need a point and a slope. Since we have two points, just pick one … IT DOESN’T MATTER … BOTH answers are acceptable… more on why later. Using the first point, we have, Using the second point, we have, Point = (6, –4) Slope = –3 Point = (2, 8) Slope = –3 f(x) + 4 = –3 (x – 6) f(x) +4 = -3x +18 f(x) = -3x +14 f(x) – 8 = –3 (x – 2) f(x) – 8 = -3x +6 f(x) = -3x +14
Other Forms of Linear Equations So far, we have discussed only point-slope form. There are other forms of equations that you should be able to identify as a line and graph if necessary. Horizontal Line: f(x) = c , where c is a constant. Example: f(x) = 3 Vertical Line: x = c , where c is a constant. Example: x = –6 Slope – Intercept Form: f(x) = mx + b m = the slope of the line … b = the y-intercept Example: f(x) = 3x – 6 Standard Form: Ax + By = C A, B, and C are integers. Example: 3x + 4y = –36
Example Rewrite each of the equations below in standard form. f(x) = x – 4 y – 6 = (x + 4)
Exit Problems Write the equation of the line that goes through the point (–3, 4) and has a slope of . 2. Write the equation of the line that passes through (2, -3) and is (a) perpendicular to and (b) parallel to the line f(x) = 2x – 3. 3. Write an equation of a line that passes through (-2, -1) and (3, 4).