10.2 Point-Slope
and Standard forms
of Linear Equations
CORD Math
Mrs. Spitz
Fall 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.
Assignment
• pgs. 408 #5-34 all
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
y2  y1
m
x2  x1
y2
2
x 1
Slope formula
Substitute values
y  2  2( x  1)
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 pointslope 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)
Point-Slope form
2
y  (4)  ( x  2) Substitute known values.
3
2
y  4  ( x  2) Simplify
3
An equation of the
line is:
2
y  4  ( x  2)
3
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 fractions
2) 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).
3
Ex. 2: Write y  4  4 ( x  2) in standard form.
3
y  4  ( x  2)
4
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.
Ex. 3: Write the standard form of an equation of
the line passing through (5, 4), -2/3
2
y  4   ( x  5)
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
Ex. 4: Write the standard form of an equation of
the line passing through (-6, -3), -1/2
1
y 3   ( x  6)
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
Ex. 6: Write the standard form of an equation of
the line passing through (5, 4), (6, 3)
y2  y1
m
x2  x1
m
First find slope of the line.
3  4 1

 1
65 1
y  4  1( x  5)
y – 4 = -1x + 5
y = -1x + 9
y+x=9
x+y=9
Substitute values and solve for m.
Put into point-slope form for conversion into
Standard Form Ax + By = C
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)
y2  y1
m
x2  x1
m
First find slope of the line.
 2 1
3 3


6  (5) 6  5 11 Substitute values and solve for m.
y 1  
3
( x  5)
11
Put into point-slope form for conversion into
Standard Form Ax + By = C
– 1) = -3(x + 5)
11y – 11 = -3x – 15
11(y
11y
11y
= -3x – 4
Multiply by 11 to get rid of fraction
Distributive property
Add 4 to both sides.
+ 3x = -4
3x + 11y = -4
Add 1x to both sides
Standard form requires x come before y.