10.2 Point-Slope and Standard forms of Linear Equations

CORD MathCORD Math

Mrs. SpitzMrs. Spitz

Fall 2006Fall 2006

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.

• pgs. 408 #5-34 all

Assignment

Application

• 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

12

12

xx

yym

21

2

x

y

)1(22 xy

Slope formula

Substitute values

Multiply each side by x-1

This linear equation is said to be in point-slope form.

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.

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)

)2(3

24 xy

An equation of the line is:

)2(3

24

)2(3

2)4(

xy

xy

Point-Slope form

Substitute known values.

Simplify

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).

Ex. 2: Write in standard form. )2(4

34 xy

)2(4

34 xy

4(y + 4) = 3(x – 2)

4y + 16 = 3x – 6)

Given

Multiply by 4 to get rid of the fraction.

Distributive property

4y = 3x – 22

4y – 3x= – 22

– 3x + 4y = – 22

3x – 4y = 22

Subtract 16 from both sides

Subtract 3x from both sides

Format x before y

Multiply by -1 in order to get a positive coefficient for x.

Ex. 3: Write the standard form of an equation of the line passing through (5, 4), -2/3

)5(3

24 xy

3(y - 4) = -2(x – 5)

3y – 12 = -2x +10

Given

Multiply by 3 to get rid of the fraction.

Distributive property

3y = -2x +22

3y + 2x= 22

2x + 3y = 22

Add 12 to both sides

Add 2x to both sides

Format x before y

Ex. 4: Write the standard form of an equation of the line passing through (-6, -3), -1/2

)6(2

13 xy

2(y +3) = -1(x +6)

2y + 6 = -1x – 6

Given

Multiply by 2 to get rid of the fraction.

Distributive property

2y = -1x – 12

2y + 1x= -12

x + 2y = -12

Subtract 6 from both sides

Subtract 1x from both sides

Format x before y

Ex. 6: Write the standard form of an equation of the line passing through (5, 4), (6, 3)

11

1

56

43

m

)5(14 xy

First find slope of the line.

12

12

xx

yym

Substitute values and solve for m.

Put into point-slope form for conversion into Standard Form Ax + By = C

y – 4 = -1x + 5

y = -1x + 9

y + x = 9

x + y = 9

Distributive property

Add 4 to both sides.

Add 1x to both sides

Standard form requires x come before y.

Ex. 7: Write the standard form of an equation of the line passing through (-5, 1), (6, -2)

11

3

56

3

)5(6

12

m

)5(11

31 xy

First find slope of the line.

12

12

xx

yym

Substitute values and solve for m.

Put into point-slope form for conversion into Standard Form Ax + By = C

11y – 11 = -3x – 15

11y = -3x – 4

11y + 3x = -4

3x + 11y = -4

Distributive property

Add 4 to both sides.

Add 1x to both sides

Standard form requires x come before y.

11(y – 1) = -3(x + 5) Multiply by 11 to get rid of fraction