Intersection of Lines

Link with Systems of Equations

The point-slope, slope-intercept, two-point and vertical line forms of the equation of a line can be written in the form 

a x + by = c 

where both coordinates x and y have been written on the left hand side of the equation and the ordered pair (a,b) is non zero, that is not (0,0).

For example

y = 5 x -10

holds when and only when

y + (-5)x = -10

It has the above form with a = 1 (not written as 1 y = y), b = -5 = (-5) and c = -10. Here the ordered pair (a, b) = (1, -5) is not (0, 0).

The equation ax + by = c when b is nonzero is equivalent to y = -(a/b)x +(c/b). The latter is the slope-intercept form of the line with slope m = -a/b and y-intercept c/b. In the case b = 0, the equation becomes ax=c or x = c/a and it provides the equation of the vertical line with x intercept c/a. So in all cases where the ordered pair (a,b) of coefficients is nonzero, the solution set of the equation ax+by= c appears as a straight line in the plane.

In general, the y intercept of the equation ax+by =c is given by y = c/b when b is nonzero; and the x intercept of the same equation is given by x = c/a when a is nonzero. And if a = 0, the equation becomes by = c, the equation for the horizontal straight line y = c/b, provided b is nonzero.

If (a,b)=(0,0) then (x,y) is a solution of the equation a x + by = c when and only when 0x+0y = c. So if c has the value 0, all point (x,y) in the plane satisfy the equation, and the equation imposes no restriction on (x,y). To avoid that equation, we assume (a,b) is nonzero.

Pairs of Lines - Geometric Expectations

Suppose L1 and L2 denote lines in the plane. These lines could be the same (concident) if unwittingly we have denoted the same line twice. These lines could be parallel or intersecting. That is what we envision geometrically.

Geometrically, if two lines L1 and L2 meet, they can only meet in one point. If the L1 and L2 meet in two points they coincide.

Pairs of Lines - The Algebraic Model

Let lines L1 and L2 be described by two equations

L1 eq'n: ax + b y = c (keep the old actors)
L2 eq'n: cx + d y = e (introduce new actors)

in which (a,b) and (c,d) are both not equal to (0,0). In brief with less clutter we write

L1: ax + b y = c
L2: cx + d y = e

A point (x,y) which satisfies both equations will belong to both lines L1 and L2 and so provide an intersection point of the lines or the solution sets for both equations.

If solving simultaneous equations is easy for you, then a large part of high school equation solving becomes very simple -- too simple, as then solutions to problems become the task, hard or not, of finding a set of linear equations to solve for the missing information in a problem.

Algebraic Example of Intersecting Lines: Find the intersection point if any of the equations

L1: 2x + 3 y = 1
L2: 5x + 9 y = 4

Solution First Part (x-elimination): Multiply the first equation by c=5 and the second equation by a=2 to get the coefficients of x in new equations to be equal. The multiplication gives the next two equations

5×L1: 10x + 15 y = 5
2×L2: 10x + 18 y = 8
2×L2-5×L1: 3 y = 3

The third equation 3y =3 follows from taking the 5×L1 equation away from the 2×L2 equation. The third equation implies y =3/3 =1. So x-elimination yields the value of y.

Solution Second Part: (Get x):

Now the first equation L1 (we could have used the second) implies 2x = 1 -3y which in turn implies the run-on set of equalities x = (1-3y)/2 = (1-3×1)/2 = (-2)/2 = -1.

So the point (x,y) = (-1, 1) belongs or should satisfy both equations and thus belong to both lines L1 and L2.

Solution Third Part (Check Answer):

Check that (x,y) = (-1, 1) satisfies the two equations

L1: 2x + 3 y = 1
L2: 5x + 9 y = 4

Whenever we obtain a solution of a set of equations, the possibility of an arithmetic or logical error in the steps that yield the solution suggests the solution should be checked. If the check fails, we correct the steps (or redo them) after checking the check to avoid looking for an error that is not in the solution.

Solution Postscripts

Uniqueness: Note the above sequence of steps that lead to the solution imply if (x,y) satisfies the two equations for L1 and L2 then (x,y) = (-1, 1). So the solution (-1,1) of the above two equations is unique. There is not a second intersection point.

Unequal Slopes: From the first of the two equations

L1: 2x + 3 y = 1
L2: 5x + 9 y = 4

we see that the slope of L1 is m1 = -2/3 while from the second of the two equations we see that the slope m2 = - 5/9 =\= -2/3 = - 6/9.

The case where the slopes are equal lead to parallel lines or coincident lines. Details follow.

The following explanations assume some knowledge of direct and indirect use of implication rules for arriving at conclusions. See the logic chapters 4, 6 and 7 in site Volume 1A, Pattern Based Reason.

NumericalExample of Parallel Lines: Find the intersection point if any of the equations

L1: -3x + 2 y = 1
L2: -9x + 6 y = 4

Solution First Part (try x-elimination)

Multiply the first equation by c=3 and the second equation by a=1 to get the coefficients of x in new equations to be equal. The multiplication gives the next two equations

3×L1: -9x + 6 y = 3
1×L2: -9x + 6 y = 4

Here we see that the left-sides of both equations are the same but the right hand side are different. So we are looking for a point (x,y) such that the expression -9x + 6 y gives two different values, namely 3 and 4, when computed. That is impossible. So there is no solution. So we are done. Our conclusion is no intersection.

Another way to see the impossibility of the two lines L1 and L2 intersecting is as follows. The first of the two equation

L1: -3x + 2 y = 1
L2: -9x + 6 y = 4

implies y = (3/2)x + 1 while the second implies y = (9/6)y + 4/6 = (3/2)x +(2/3). Now we have two new equations for L1 and L2, namely

L1: y = (3/2)x + 1
L2: y = (3/2)x + 2/3

The slope-intersection form of these equations tells us that both lines have the same slope 3/2, the y-intercept of the first is 1 while the while intercept of the second is 2/3. We see for each point x on the x-axis, the y coordinate of a point (x,y1) on the first line L1 is 1/3 more than the y-coordinate of a point (x,y1) on the second line L2. See the diagram

The intersection of a vertical line with L1 (the blue line) is 1/3 of a unit above its intersection with L2 (the red line). So there is no intersection. (An intersection point (p, q) would give a vertical line x = p in which the intersection with L1 and L2 have the same height q, but the intersections with L1 and L2 of x=p always have different heights.

Algebraic Example of Coincident Lines: Find the intersection point if any of the equations

L1: 3x + 9 y = 6
L2: 5x + 15 y = 10

Solution First Part (try x-elimination): Multiply the first equation by c=5 and the second equation by a=5 to get the coefficients of x in new equations to be equal. The multiplication gives the next two equations

5×L1: 15x + 45 y = 30
3×L2: 15x + 45 y = 30

Here we see that the left-sides and rights of the resulting equations are the same. So we are looking for a point (x,y) such that the expression 15x + 45 y gives the value 30. Here they are many solutions, the set or line described by the equation

15x + 45 y = 30

Now let us retreat and rewrite each equation for L1 and L2 in slope-intercept form.

• The L1 equation 3x + 9 y = 6 gives y = (-3x + 6)/9 or y = -(1/3) x + 2/3
• The L2 equation : 5x + 15 y = 10 gives y = (-5x +10)/15 or equivalently y = -(1/3) x + 2/3

So L1 and L2 equations both represent the line y = -(1/3) x + 2/3. All points on this line satisfy the L1 and L2 equations. That implies L1 and L2 denote the same line.

Arithmetic - Ages 10+
1. Deciml Place Value - fun
2. Decimals for Tutors
3. Prime Factors - quickly
4. Fractions + Ratios
5. Arith with units - science