Ch. 10 The Rectangular Coordinate System and Linear
Equations
ordered pair (x,y):
The first number in the ordered
pair is the x-coordinate and the
second number in the ordered pair
is the y-coordinate.
The line y = - ½ x + 1 is the set of
ordered pairs (x,y) for which that
equation is true.
x- intercept - point where graph
crosses x-axis. y = 0 at this point.
y-intercept – point where graph
crosses y-axis. x = 0 at this
point.
slope = rise/run of a line, aka
If two points of a line are known, (x1,y1) and (x2,y2), the slope can be found with this
equation, the slope,
m
( y 2  y1 )
( x2  x1 )

change in y
change in x
3 points are “collinear” if each pair has the same slope.
( y  y1 ) ( y3  y1 ) ( y3  y2 )
Example: (x1,y1), (x2,y2) and (x3,y3) are collinear if m  2


( x2  x1 )
( x3  x1 )
( x3  x2 )
2 lines are parallel if they have the same slope (m1 = m2), and they are perpendicular have
their slopes are opposite reciprocals (m1 = -1/m2)
Example: If one line has a slope of 2/3, then all lines parallel to it would also have a slope
of 2/3 and all lines perpendicular to it would have a lope of -3/2.
FINDING THE DISTANCE BETWEEN TWO POINTS
The distance, d, between two points can be found using the Pythagorean Theorem.
Find the distance d between points (1,3) and (5,6)
Solution: Create a right triangle by drawing a horizontal line from (1,3) to (5,3) [ since 5
is the x-coordinate of (5,6), and a vertical line from (5,3) to (5,6). The line segment from
(1,3) to (5,6) is the hypotenuse, c. The other line segments are the legs, a and b.
The distance between two points on a horizontal line is just the absolute value of the
difference of their x-coordinates (|x2 – x1|), so the length of a = |5 – 1| = 4. Likewise,
the distance between two points on a vertical line is just the absolute value of the
difference of their y-coordinates (|y2 – y1|), so the length of b = |6 – 3| = 3
From the Pythagorean Theorem, c2 = a2 + b2 = 42+32 = 16+9 = 25
25 = 5
c=
7
(5,6)
6
5
b
c
4
3
a
(1,3)
(5,3)
2
1
0
0
1
2
3
4
5
DISTANCE FORMULA
The distance between two points P1 = (x1,y1) and P2 = (x2,y2) is
d(P1,P2) =
( x2  x1 ) 2  ( y2  y1 ) 2
6
Using Algebra to Solve Geometry Problems
Consider three points A = (-2,1), B = (2,3), and C = (3,1)
a) Plot each point and form triangle ABC.
5
4
B = (2,3)
y
3
2
1
C = (3,1)
A = (-2,1)
0
-3
-2
-1
0
1
2
3
4
x
b) Find the length of the three sides of the triangle, AB, BC, and AC
d (A,B) =
d (B,C) =
d(A,C) =
c) Is ΔABC a right triangle?
d) Find the area of the triangle.
The area of a triangle is ½ (base)(height)
If you rotated the triangle to lay on one of it’s legs instead of the hypotenuse, you can see that
the base is AB and the height is BC (or vice versa).
So Area =
Midpoint Formula
The midpoint M = (x,y) of the line segment from
P1 = (x1,y1) to P2 = (x1,y2) is
 x1  x2 y1  y2 
M  ( x, y)  
,

2 
 2
Therefore, to find the midpoint of a line segment, we average the xcoordinates and the y-coordinates of the endpoints.
Examples:
a) Find the midpoint of this line segment.
Two points have “symmetry” if they are equidistant from a
line or point. If 2 points have symmetry with respect to a point.
Then are equidistant from the point and collinear with the
point.
5
4
3
2
1
0
-5
-4
-3
-2
-1
0
-1
-2
-3
-4
1
2
3
4
5
5
4
3
2
1
0
-1 0
-2
-3
-4
-5
1
2
3
4
5
6
7
5
4
3
2
1
0
-7 -6 -5 -4 -3 -2 -1 0
1
2
3
4
5
6
7
Example: Show that the following 4 ordered pairs form an isosceles triangle.
Definition of isosceles trapezoid –
A quadrilateral with one pair of opposite
sides being parallel and the other pair of
opposites being congruent (same length).
(3,8)
(1,4)
(9,8)
(11,4)
EQUATIONS OF STRAIGHT LINES
Example 1:
Find the equation of the line that has slope 3 and y-intercept (0,2)
This is easy because the slope-intercept form of a line,
y=mx +b, has m=slope, and b = the y-coordinate of the y-intercept. So
plugging in m=3 and b=2, we get
y = 3x + 2
Example 2: Find the equation of the line that passes throught the point
whose coordinates (-2,-1) and slope = 3/2.
USE THE POINT-SLOPE FORMULA.
Recall that the slope, m, can be found by this equation:
y2  y1
x2  x1
If we want an equation, we will not plug in values
for one of the points (x 2 , y 2 ), instead we will just use the variables (x, y)
y  y1
m
x  x1
Now for an equation with no fractions,
multiply both sides by x  x1 
y  y1
x  x1 
mx  x1  
x  x1
mx  x1   y  y1
The POINT - SLOPE FORMULA puts the y-y1 on the left, so we get
y  y1  mx  x1 
m
m=
FIND THE EQUATION OF A LINE GIVEN TWO POINTS
We need a point and a slope to use the point-slope formula.
So this type of problem requires 2 steps:
Step 1) Find the slope using the two points and the formula:
m
y2  y1
x2  x1
Step 2) Use one of the given points and the slope you just found in the
Point-slope formula and then get y by itself.
y  y1  mx  x1 
Example 3: Find the equation of the line that passes through the points
whose coordinates are (-4,-5) and (8,4)
Step 1) Find the slope. Let (x1,y1) = (-4,-5) and (x2,y2)=(8,4)
Step 2) Use the point-slope formula and get y by itself.
FIND THE EQUATION OF A LINE GIVEN THAT IT IS
PARALLEL OR PERPENDICULAR TO ANOTHER
LINE.
First, know these facts:
A. Parallel lines have the same slope.
B. The slopes of Perpendicular lines are negative reciprocals.
Second, use the same steps as before.
Step 1) Find the Slope of the line you are finding.
Step 2) Use the Point-Slope Formula and get y by itself to
make an equation.
Example:
Write an equation for the line that is parallel to 4x + y = -9 and
contains the point (-2,9)
Step 1) Find the slope
Step 2) Use the Point-Slope Formula
Example:
Write an equation for the line that is perpendicular to the line,
3x + 2y = -10 that contains the point (-9,-2)
Step 1) Find the Slope
Lines that are perpendicular have slopes that are opposite reciprocals.
What’s the slope of the first line?
Step 2) Use the Point-Slope Formula
Example:
Write an equation for the perpendicular bisector that goes through
the line segment with endpoints A (0,1) and B (4,9).
The perpendicular bisector goes through the midpoint of
AB and is perpendicular to
AB
Step 1) Find the slope of
AB
m = Slope =
Step 2) Take the negative reciprocal of the slope
Slope perpendicular to _______ is ____
Step 3) Find the midpoint of
AB
Step 4) Use the point-slope formula to find the equation with
m = ___and (x1,y1) = _________