This article assumes a knowledge of basic algebra.
SLOPE OF A STRAIGHT LINE
The SLOPE OF A STRAIGHT LINE like the one shown in figure one is
m = (y2 – y1)/(x2 – x1) Equation one
where the line extends from (y2,x2) to (y1,x1).
EQUATION FOR A STRAIGHT LINE
The general algebraic expression for the line is
y = m*x + b Equation two
To define the constant b, we set x equal to zero.
y = m*0 + b
y = b
Hence when x is equal to zero, y is equal to b
TWO PARALLEL LINES
If we have two parallel lines as shown in figure two, then the slop of each line is the same.
The slope of one line will be designated m1 and the slope of the other line will be designated as m2.
y = m1*x + b Equation three
y = m2*x + c Equation four
m1 = m2 for two parallel lines.
TWO PERPENDICULAR LINES
If two lines are perpendicular to each other, then the slope of one line is the negative reciprocal of the slope of the other line. See figure three.
If m1 is the slope of one line, the slope m2 of the perpendicular line is
m2 = -1/m1
TWO LINES NOT PARALLEL WITH EACH OTHER
If the two lines are not in parallel with each other, then the slope of the first line m1 will not equal the slope of the second line m2
y = m1*x + b
y = m2*x + c
m1 m2
If these two lines are extended in both directions, at some point they will intersect each other. To find out where they intersect each other, we set up simultaneous equations and solve to y and x.
PROBLEMS AND SOLUTIONS
The two points (x,y) defining line one are (3,5) and 4,6)
What is the slope m1 of line one?
m1 = (y2 – y1)/(x2 – x1) = (6 – 5)/(4 – 3)
m1 = 1/1 = 1
The two points (x,y) defining line two are (5,6) and (7,9)
What is the slope m2 of line two?
m2 = (y2 – y1)/(x2 – x1) = (9 – 6)/(7 – 5)
m2 = 3/2 = 1.5
Find the value of b in the equation y = m1*x + b
m1 = 1
The value of b is equal to the value of y with x equal to 0.
We use the equation for the slope to figure out the value of y when x equals 0.
m1 = (y2 – y1)/(x2 – x1) = (y1 – y0)/(x1 – x0) Equation five
where
y0 = b
x0 = 0
Substituting values into equation one, we get
(6 – 5)/(4 – 3) = (5 – b)/(3 – 0)
1/1 = (5 – b)/(3 – 0)
3 – 0 = 5 – b
3 = 5 – b
b = 2
Hence,
The equation becomes
y = m1*x + 2
or
y = 1*x + 2
Find the value of c in the equation y = m2*x + c
y = m2*x + c
m2 = 3/2
y = (3/2)*x + c
m2 = (y2 – y1)/(x2 – x1) = (y1 – c)/(x1 – 0)
m2 = (9 – 6)/(7 – 5) = (6 – c)/(5 – 0)
3/2 = (6 – c)/5
5*(3/2) = 6 – c
15/2 – 6 = -c
c = 6 – 15/2
c = 12/2 – 15/2
c = -3/2
Hence our equation becomes
y = (3/2)*x – 3/2
Find the point at which the following lines intersect
y = 1*x + 2 Equation six
y = (3/2)*x – 3/2 Equation seven
Converting the whole numbers into fractions in equation six, we get
y = (2/2)*x + 4/2
y = (3/2)*x – 3/2
From equation six
x = y – 2
Substituting this into equation seven we get
y = (3/2)*(y – 2) – 3/2
Solving for y
y = (3/2)*y – 6/2 – 3/2
y – (3/2)*y = -9/2
-(1/2)*y = -9/2
(1/2)*y = 9/2
y = 18/2 = 9
if y = 9 and y = x + 2 then
x = 7
Intersection point is y = 9 and x = 7
Proof
y – (3/2)*y = -9/2
9 = (3/2)*7 – 3/2
9 = 21/2 – 3/2
9 = 18/2
9 = 9
The coordinates (9,7) fits both equation six and equation seven and therefore exists on both lines.
Plot the two lines defined by the following equations and show their intersection points on the plot.
y = 1*x + 2 Equation six
y = (3/2)*x – 3/2 Equation seven
Equation six yields the following points
(y,x) (9,7) (6,4) (5,3)
Equation seven yields the following points
(y,x) (9,7) (6,5)
See figure four for the plot and the intersecting point.
References:
New Second Algebra
Library of Congress Catalog Card Number 62-7240