Chapter 2
The Straight Line and the Circle
2.1 Lines and First-Degree
Equations
Page 51
Objectives:
1. Finding the equation of a line given
slope and y-intercept, given a point
and slope, and given two points.
Theorem 2.1
• The equation of every straight line is
expressible in terms of the first degree.
Conversely, the graph of a first-degree
equation is a straight line.
Example 1 Write the equation of the
line with slope -3 and y-intercept 4.
Graph the line.
Example 2 Express the equation 4x –
3y – 11 = 0 in the slope-intercept form
and graph.
Slope Intercept Form
Y = mx + b
Point-Slope Form
Y – Y1 = m(X – X1)
Example 3 Find the equation of the
line that passes through (2, -3) and has
a slope of 5.
Example 4 Find the equation of the line
determined by the points (3, -3) and (2, 4).
Find the X-Intercept and YIntercept of 8x – 6y = 192
Find the Slope of the Line
4x – 5y = - 5.
Find the slope of the line that goes
through (1,-5) and (-4,-3).
Find the Equation of the line
Parallel to the line: y = -3x + 1 and
going through (2,-4)
Find the Slope of the line
Perpendicular to the line: 2x – 2y = 3.
Find the equation of the line
perpendicular to the line: -3x – 4y =
-3 and going through (-3,4).
Write the Equation of the line
having y-intercept: 1; and slope: 4.
WRITE THE EQUATION OF THE LINE
Describing:
X
Y
-2
-3
-4
-5
-6
-7
-8
-9
Find the Equation of the line
parallel to the line: y = -3x – 5 going
through (2,-3).
Write the equation in standard
form: y = -3x – 3.
Write the equation in slopeintercept form: y + 3 = 3(x + 2).
Homework Assignment
Page 57
Problems:
4, 12, 19, 31, 42, 51, 53, 59.
2.2 Other Forms Of First-Degree
Equations
Page 59
Objectives:
1. Find the equation of a line given slope and
passing through a point, given x-intercept and
y-intercept, and given a point and equation
that is perpendicular to the line.
Alternative form for the point-slope
equation.
To obtain this form, we substitute –(A/B) for m
in y – y1 = m(x – x1).
A
y  y1    x  x1 
B
Multiplying by B and transposing terms yields
Ax  By  Ax1  By1
Intercept Form
Suppose the x-intercept of a line is a and the y-intercept is b, where a  0 and b  0.
b
Then the line passes through the point (a,0) and (b,0), and the slope is -   .
a
Then, applying the point-slope formula, we find the equation
b
 x  0 ,
a
which can be reduced to
y b  
x y
  1.
a b
Intercept Form
b
y  b    x  0
a
a  y  b   bx
ay  ab  bx
bx  ay  ab
bx ay ab


ab ab ab
x y
 1
a b
Example 2 Write the equation of the
line whose x-intercept is 3 and whose
y-intercept is -5.
x y

 1 or 5 x  3 y  15
3 5
Example 3 Write the equation 4x – 9y
= -36 in the intercept form.
Divide the given equation by -36
x
y
 1
9 4
Example 4 Find the equation of the line through the
point (6,2) and perpendicular to the line defined by the
equation 4x + 5y + 7 = 0.
Let 5x – 4y be the
left side of our
equation.
Apply Ax  By  Ax1  By1
With x1 = 6
y1 = 2
5x -4y =5(6) – 4(2)
Or 5x – 4y = 22
Example 5 The ends of a line segment are at
C(7,-2) and D(1,6). Find the equation of the
perpendicular bisector of the segment CD.
Definition 2.1
Definition 2.2
Theorem 2.2
HOMEWORK ASSIGNMENT
Page 63: 2, 11, 17,
8, 25, 29,37,