The general equation of a line in a 2 dimensional x-y plane may be written as
\[ a x + b y = c \]
The coefficients \( a, b \) and \( c \) determine the properties of the line and gives information on whether the line is horizontal or vertical or neither (slanted); and also the x and y intercepts of the line.
Example 1
a) Find four ordered pairs that are solutions to the linear equation in two variables given by: \( \quad y + 2x = 1 \)
b) Plot the ordered pairs found in part a) above and join the points obtained.
Solution to Example 1
a)
Because the given equation \( y + 2 x = 1 \) has two variables \( x \) and \( y \), it has an infinite number of ordered pairs solutions \( (x,y) \) that may be found by giving a value to one of the two variables and find the value of the second variable.
1) let \( x = 2 \) and substitute it in the given equation \( y + 2x = 1 \) to obtain \( y + 2 \times 2 = 1 \)
Simplify the left side of the equation
\( y + 4 = 1 \)
Solve for \( y\) : \( y = 1 - 4 = - 3\), hence the ordered pair \( \color{red}{(2,-3)} \) is a solution to the equation \( y + 2x = 1 \) .
2) let \( x = -1 \) and substitute in the given equation \( y + 2x = 1 \) to obtain \( y + 2 \times (-1) = 1 \)
Simplify the left side of the equation
\( y - 2 = 1 \)
Solve for \( y\) : \( y = 1 + 2 = 3\), hence the ordered pair \( \color{red}{(-1,3)} \) is a solution to the equation \( y + 2x = 1 \)
3) let \( y = 5\) and substitute in the given equation \( y + 2x = 1 \) to obtain \( 5 + 2 x = 1 \)
Solve for \( x\) : \( 2x = 1 - 5 = - 4\) which gives \( x = - 2 \) , hence the ordered pair \( \color{red}{(-2,5)} \) is a solution to the equation \( y + 2x = 1 \)
4) let \( y = -1\) and substitute in the given equation \( y + 2x = 1 \) to obtain \( -1 + 2 x = 1 \)
Solve for \( x\) : \( 2x = 1 - (-1) = 2\) which gives \( x = 1 \) , hence the ordered pair \( \color{red}{(1,-1)} \) is a solution to the equation \( y + 2x = 1 \)
We may continue and find as many ordered pairs as we want. Here are some more ordered pairs that are solutions to the equation \( y + 2x = 1 \):
\( \color{red}{(0,1) , (1/2 , 0) , (3 , - 5) , (-1/2 , 2) ....} \)
Note that you may check that all the ordered pairs obtained above are solutions to the given equation by substitution.
b)
We now plot the ordered pairs found in part a) and join the points obtained. The graph obtained by joining the points is a slanted line.
Conclusion
The graph of any equation of the form \( \; ax + by = c \; \) is a line.
Horizontal Lines : \( a = 0 \)
If the coefficient \( a \) in the linear equation \( ax + by = c \) is equal to zero, the equation becomes:
\( 0 x + b y = c \) or \( b y = c \)
Divide all terms of the equation by \( b \) to obtain
\( y = c / b = B\)
The ordered pairs that are solutions to the above equation \( y = B\) are of the form
\[ (k , B) \]
where \( k \) is any real number.
The ordered pairs \( (k , B) \) form a horizontal line. Hence the graph of an equation of the form \( y = B \) is a horizontal line.
Example 2
a) Find four ordered pairs that are solutions to the linear equation \( \quad y = 3 \) of a line in 2 dimensional plane.
b) Plot the ordered pairs found in part a) above and join the points obtained.
c) Give two more examples of horizontal lines
Solution to Example 2
a)
The given equation of a line in a 2 dimensional plane may be written as
\[ 0 x + \quad y = 3 \]
Hence, any ordered pair of the form \( (k , 3) \) is a solution to the given equation \( \quad y = 3 \)
Some ordered pairs that are solutions: \( (0, 3) , (-2,3) , (3,3) , (4,3) , (-4,3) ...\)
b) The ordered pairs listed in part a) are plotted in the graph below and joined. The graph obtained by joining the points is horizontal line that passes through all points with y coordinate equal to 3.
c) More examples of horizontal lines:
\( y = -3 \) this is the equation of a horizontal line that passes through all points with y coordinate equal to \( -3 \).
\( y = 0 \) this is the equation of a horizontal line that passes through all points with y coordinate equal to \( 0 \) (x axis).
Vertical Lines : \( b = 0 \)
If the coefficient \( b \) in the linear equation \( ax + by = c \) is equal to zero, the equation becomes:
\( a x + 0 y = c \) or \( a x = c \)
Divide all terms of the equation by \( a \) to obtain
\( x = c / a = A\)
The ordered pairs that are solutions to the above equation \( x = A\) are of the form
\[ (A , k) \]
where \( k \) is any real number.
The ordered pairs \( (A , k) \) form a vertical line. Hence the graph of an equation of the form \( y = A\) is a vertical line.
Example 3
a) Find four ordered pairs that are solutions to the linear equation \( \quad x = - 2 \) of a line in 2 dimensional plane.
b) Plot the ordered pairs found in part a) above and join the points obtained.
c) Give two more examples of horizontal lines
Solution to Example 3
a)
The given equation of a line in a 2 dimensional plane may be written as
\[ x + 0 y = - 2 \]
Hence, any ordered pair of the form \( (-2 , k) \) is a solution to the given equation \( \quad x = - 2 \)
Some ordered pairs that are solutions: \( (-2, 3) , (-2,-3) , (-2,1) , (-2,0) , (-2,-1) ...\)
b) The ordered pairs listed in part a) are plotted in the graph below and joined. The graph obtained by joining the points is vertical line that passes through all points with x coordinate equal to -2.
c) More examples of vertical lines:
\( x = 3 \) this is the equation of a vertical line that passes through all points with x coordinate equal to 3.
\( x = 0 \) this is the equation of a vertical line that passes through all points with x coordinate equal to 0 (y axis).
x and y Intercepts of the Graph of a Line
We now explore the x and y intercepts of the general equations of lines with equations
\[ a x + b y = c \]
where neither \( a \) nor \( b \) is equal to zero.
The \( x \) intercept is found by setting \( y = 0 \) in the above equation and solve for \( x \).
\( ax + b(0) = c \)
Solve the above equation for \( x \) to obtain
\( x = c/a \)
Hence, the \( x \) intercept is at the point \( (c / a , 0) \).
The y intercept is found by setting \( x = 0 \) in the above equation and solve for x.
\( a(0) + by = c \)
Solve the above equation for \( y \) to obtain
\( y = c/b \)
Hence, the \( y \) intercept is at the point \( (0 , c/b) \).
Example 4
Find the x and y intercepts of the graph of the equation given by
a) \( 2 x - y = 2 \)
b) \( 4 x + 2 y = 0 \)
Solution to Example 4
a) The x intercept of \( 2 x - y = 2 \) is found by setting \( y \) to 0.
\( 2 x - 0 = 2 \)
Solve \( 2 x = 2 \), which gives \( x = 1 \).
The x intercept is at \( (1 , 0) \).
The y intercept of \( 2 x - y = 2 \) is found by setting \( x \) to 0.
\( 2 (0) - y = 2 \)
Solve\( - y = 2 \) , which gives \( y = - 2 \).
The y intercept is at \( (0 , -2) \).
b) The x intercept is found by setting \( y = 0 \) and solving \( - 4 x = 0 \) , which gives \( x = 0 \).
The x intercept is at (0 , 0).
The y intercept found by setting \( x = 0 \) and solving \( 2 y = 0 \), which gives \( y = 0 \).
The y intercept is at \( (0 , 0) \).
In fact any equation of the form \( a x + b y = 0 \) has x and y intercepts at the origin \( (0,0) \).
