TOPIC 4.1: GRAPHS
4.1.1: The Distance and Midpoint
Formulas
4.1.2: Graphs of Equations in Two
Variables; Intercepts; Symmetry
4.1.3: Lines
4.1.4: Circles
4.1.1: The Distance and Midpoint
Formulas
Rectangular Coordinate System
Horizontal axis: x-axis ; Vertical axis
: y-axis
Origin O: the point of intersection of the two axes.
Coordinates - location of a point P(x,y) (an ordered pair) on the xy-plane.
y
Quadrant II
x < 0, y > 0
Quadrant I
x > 0, y > 0
x
Quadrant III
x < 0, y < 0
Quadrant IV
x > 0, y < 0
x axis
origin
Rectangular or Cartesian Coordinate System
Distance Between Points
The distance d between two points (x1, y1 ) and (x2 , y2 )
in the rectangular coordinate system is:
d = ( x2 − x1 ) 2 + ( y2 − y1 ) 2
Example:
Find the distance between (2, -2) and (5, 2)
Midpoint Formula
The midpoint of a line segment from (x1 , y1 ) to (x 2 , y 2 ) is
 x + x 2 y1 + y 2 
( x, y ) =  1
,

2
2


Example:
Find the midpoint of the line segment with endpoints (1, 2) and (7, -3)
4.1.2: Graphs of Equations in Two
Variables; Intercepts; Symmetry
The graph of an equation in two variables
x and y consists of the set of points in the
x-y plane whose coordinates (x,y) satisfy
the equation
Intercepts
-The points at which a graph intersects the coordinate axes.
i) x-intercept - crosses/touches the x-axis.
ii) y-intercept - crosses/touches the y-axis.
1.
2.
To find the x-intercept, substitute y = 0 in the equation and solve for x.
To find the y-intercept, substitute x = 0 in the equation and solve for y.
Exercise:
Find x-intercept(s) and y-intercept(s) for y = x2 - 9
Find the x-intercept(s) and the y-intercept(s)
2
of the graph of y = x − 4 then graph by
plotting points.
Symmetry
Test for Symmetry: y = f(x)
- with respect to the x-axis: replace y by (-y) in the equation.
- with respect to the y-axis: replace x by (-x),
- with respect to the origin: replace x by (-x) and y by (-y)
Exercise:
Test the graph of x = y2 for symmetry with respect to x-axis
4.1.3: Lines
Definition of slope:
The slope of the line through the distinct points (x1, y1) and (x2, y2) is
m =
y 2 − y1
,
x 2 − x1
x 2 ≠ x1
Example: Find the slope of the line passing through each pair of points:
a) (-3, -1) and (-2, -4)
b) (4, -2) and (-1, 5)
Positive slope (m > 0)
Zero slope (m = 0)
Negative slope (m < 0)
Undefined slope (m = ∞ )
Equations of lines
1. Point-slope form:
y − y1 = m( x − x1 )
2. Slope-intercept form:
y = mx
+ c
3. Horizontal line:
y
=
b
4. Vertical line:
x = a
5. General form:
Ax + By + C = 0
Example:
1. Write the point-slope equation of the line that passes through (-1, 3)
with slope 6. Then solve for y
3
x + 1
2. Graph the line whose equation is y =
5
Tips;
1.plot y-intercept (0,b)
2.Obtain a second point by using the formula of gradient, m.
3.Use a straightedge to draw a line through the 2 points
3. Graph x = 5 and y = 2
4. Find the slope and y-intercept of 2x – 3y + 6 = 0
Parallel and Perpendicular Lines
Let two lines L1 : y = m1 x + c1 and L2 : y = m2 x + c 2 , and
L1 and L2 are parallel ⇔ m1 = m2
L1 and L 2 are perpendicular
⇔
m1 ⋅ m2 = −1
Example:
1. Write an equation of the line that passes through (-2,5) and parallel to the line
whose equation is y= 3x +1. Express the equation in slope intercept form
2. Find the slope of any line that is perpendicular to the line whose equation is
x+ 4y – 8 = 0. Then, write it in slope intercept form
L1 : − 3 x + 2 y = 12
L2 : 6 x − 4 y = 0
2 y = 3x + 12
3
y = x+6
2
3
Slope = ; y -intercept = 6
2
−4 y = −6 x
3
y= x
2
3
Slope = ; y -intercept = 0
2
y
6
5
4
3
2
1
x
−6
−5
−4
−3
−2
−1
1
−1
−2
2
3
4.1.4: Circles
A circle is a set of points in the xy-plane that are equidistant from a fixed point,
called the center. The fixed distance from the circle’s center to any point on
the circle is called the radius.
The standard form of an equation of a circle with radius r and center (h,k):
( x − h) 2 + ( y − k ) 2 = r 2
If the center is at the origin (0,0)
⇒
x2 + y2 = r 2
If the radius is 1, (r = 1)
⇒ x2 + y2 =1
(unit circle)
The general form of the equation of a circle is:
x 2 + y 2 + ax + by + c = 0
Example:
1-Write the standard form of the equation of the circle with center (5, -6) and radius 10
2- Find the center and radius of the circle whose equation is ( x − 2) 2 + ( y + 4) 2 = 9
3- Write in standard form and sketch:
x2 + y2 + 4x − 4 y −1 = 0