Thus the coördinates of any point on the line satisfy the equation.

Second. Any point whose coördinates satisfy the equation lies on the line.

Let the coördinates (x', y') of the point P (Figure 2) satisfy the equation. Then we have

or

y' = ax',

Let the ordinate y' cut the line at B. Then by the first part of the proof

BC ax',

BC

=

Hence P lies on the line.

THEOREM II. The graph of any linear equation in two vari

the equation Ax + C = 0 may be put in the form

C

provided A0. This is evidently the equation of a straight line parallel to the Y axis (§ 102). If B = 0 and A = 0, we have no term left involving the variable. Thus the only case for which the theorem demands proof is when B 0, and the equation may be reduced to form (2). By Theorem I we know that the graph of y = ax is a straight line. If, then, we add to

y= ax+b

ах

y = ax

every ordinate У of the line y = ax the constant b, the locus of the extremities of the lengthened ordinates, will lie in a straight line, as one can easily prove by Geometry. But any point (x, y) on the upper line is such that its ordinate y is equal to the ordinate of the lower line, i.e. ax, and in addition the constant b; that is, y = ax + b. Also, since the upper line is the locus of the extremities of the lengthened ordinates, every point whose coördinates satisfy the equation y = ax + b is on this upper line. Thus the equation (1) has a straight line as its graph.

COROLLARY. Two lines whose equations are in the form

y = ax + b,

y = = ax + b'

(3)

are parallel.

(4)

For the value of the ordinates of (3) corresponding to a given abscissa, say a1, is obtained from the ordinate of (4) corresponding to the same abscissa by adding the constant bb'. Thus each point on (3) is always found b-b' units above (below if bb' is negative) a point of (4). Thus the lines are parallel.

106. Method of plotting a line from its equation. Since the equation y = ax + b is satisfied by the values (0, 6), the graph cuts the y axis b units above (below if b is negative) the origin. Since it is satisfied by the values (1, a + b), the graph passes through the point reached by going one unit of r to the right of (0, b) and a units up (down if a is negative). These two points determine the line. We may then plot a linear equation by the following

RULE. Reduce the given equation to the form

y = = ax + b.

Plot the point (0, b) as one of the two points that determine the line.

From this point go one unit of x to the right and a units of y up (down if a is negative) to find a second point that lies on the line.

Draw the line through these two points.

0 (1,-1)

From this point go one unit of x to the right and three units of y down to find the second point, which helps

107. Solution of linear equations, and the intersection of their graphs. The process of solving a pair of independent linear equations consists in finding a pair of numbers (x, y) which satisfy them both. Though each equation alone is satisfied by countless pairs of values (x, y), we have seen that there is only one pair that satisfy both equations. Since a pair of values which

satisfy an equation are the coördinates of a point on its graph, it appears that the pair of values that satisfy simultaneously two equations are the coördinates of the point common to the graphs of the two equations, that is, the coördinates of the points of intersection of the two lines.

EXERCISES

Find the solutions of the following equations algebraically. Verify the results by plotting and noting the coördinates of the point of intersection.

YA

44-239'3

To plot (1) and (2) we get the equations in the form y = ax + b and apply

the rule. Thus

108. Graphs of dependent equations. We have defined (§ 57) dependent equations as those that are reducible to the same form on multiplying or dividing by a constant. Thus two dependent equations are reducible to the same equation of the form y = ax + b. Hence dependent equations have as their graphs the same straight line. We see now the geometrical meaning of the statement that dependent equations have countless common solutions. Since their graphs have not one but countless points in common, being the same line, it is clear that the coördinates of these countless points will satisfy both equations.

109. Incompatible equations. By our definition (§ 60) incompatible equations have no common solution. Since every pair of distinct lines have a common point unless they are parallel, we can foresee the

THEOREM. Incompatible equations have parallel lines as graphs.

Let the equations

ax + by + c = 0,

(1)

a'x + b'y + c′ = 0

(2)

be incompatible. This is true (§ 60) when and only when

This may be done if neither b nor b' equals zero. If both b and b' vanish, the lines (1) and (2) are both parallel to the Y axis and hence to each other, which was to be proved. But if only one of them vanishes, say b = 0, then by (3) a = 0 (§ 5), in which case (1) does not include either variable. Thus we may assume that neither b nor b' vanishes and that (4) and (5) may be obtained from (1) and (2).

which represent parallel lines, by the Corollary, p. 96.

This theorem completes the discussion of the graphical representation of the possible classes (§ 61) of pairs of linear equations.