10.2 Point-Slope and Standard forms of Linear Equations

1. 10.2 Point-Slope and Standard forms of Linear Equations CORD Math Mrs. Spitz Fall 2006

2. Objective • Write a linear equation in standard form given the coordinates of a point on the line and the slope of the line. • Write a linear equation in standard form given the coordinates of two points on a line. Assignment • pgs. 408 #5-34 all

3. Seth is reading a book for a book report. He decides to avoid a last minute rush by reading 2 chapters each day. A graph representing his plan is shown at the right. By the end of the first day, Seth should have read 2 chapters, so one point on the graph has coordinates of (1, 2). Since he plans to read 2 chapters in 1 day, the slope is 2/1 or 2. Application

4. Application Slope formula Substitute values Multiply each side by x-1 This linear equation is said to be in point-slope form.

5. Point-Slope Form • For a given point (x1, y1) on a non-vertical line with slope m, the point-slope form of a linear equation is as follows: y – y1 = m(x – x1) In general, you can write an equation in point-slope form for the graph of any non-vertical line. If you know the slope of a line and the coordinates of one point on the line, you can write an equation of the line.

6. Ex. 1: Write the point-slope form of an equation of the line passing through (2, -4) and having a slope of 2/3. y – y1 = m(x – x1) Point-Slope form Substitute known values. Simplify An equation of the line is:

7. Standard Form • Any linear equation can be expressed in the form Ax + By = C where A, B, and C are integers and A and B are not both zero. This is called standard form. An equation that is written in point-slope form can be written in standard form. • Rules for Standard Form: • Standard form is Ax + By = C, with the following conditions:1) No fractions2) A is not negative (it can be zero, but it can't be negative).By the way, "integer" means no fractions, no decimals. Just clean whole numbers (or their negatives).

8. Ex. 2: Write in standard form. Given 4(y + 4) = 3(x – 2) Multiply by 4 to get rid of the fraction. 4y + 16 = 3x – 6) Distributive property 4y = 3x – 22 Subtract 16 from both sides 4y – 3x= – 22 Subtract 3x from both sides – 3x + 4y = – 22 Format x before y 3x – 4y = 22 Multiply by -1 in order to get a positive coefficient for x.

9. Ex. 3: Write the standard form of an equation of the line passing through (5, 4), -2/3 Given 3(y - 4) = -2(x – 5) Multiply by 3 to get rid of the fraction. 3y – 12 = -2x +10 Distributive property 3y = -2x +22 Add 12 to both sides 3y + 2x= 22 Add 2x to both sides 2x + 3y = 22 Format x before y

10. Ex. 4: Write the standard form of an equation of the line passing through (-6, -3), -1/2 Given 2(y +3) = -1(x +6) Multiply by 2 to get rid of the fraction. 2y + 6 = -1x – 6 Distributive property 2y = -1x – 12 Subtract 6 from both sides 2y + 1x= -12 Subtract 1x from both sides x + 2y = -12 Format x before y

11. Ex. 6: Write the standard form of an equation of the line passing through (5, 4), (6, 3) First find slope of the line. Substitute values and solve for m. Put into point-slope form for conversion into Standard Form Ax + By = C y – 4 = -1x + 5 Distributive property y = -1x + 9 Add 4 to both sides. y + x = 9 Add 1x to both sides x + y = 9 Standard form requires x come before y.

12. Ex. 7: Write the standard form of an equation of the line passing through (-5, 1), (6, -2) First find slope of the line. Substitute values and solve for m. Put into point-slope form for conversion into Standard Form Ax + By = C 11(y – 1) = -3(x + 5) Multiply by 11 to get rid of fraction 11y – 11 = -3x – 15 Distributive property 11y = -3x – 4 Add 4 to both sides. 11y + 3x = -4 Add 1x to both sides 3x + 11y = -4 Standard form requires x come before y.