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# Intersection of Two Lines

When the gradients of two straight lines are not equal or the product of their gradients is not equal to -1, then these two lines are not parallel. In that situation, it will result in the intersection of two lines.

To find the point of intersection of two lines, we solve the simultaneous equations.

Example 1: Find the point of intersection of 2x + 3y = 5 and 3x – 4y = 6

2x + 3y = 5    A

3x – 4y = 6     B

(A x 3) – (B x 2) gives

 6x + 9y = 15 – 6x + 8y = -12 17y = 3

So y =

Substitute y = in A

2x + (3 x )  =  5

2x  =  5 –

2x  =

2x  =

x  =

So the point of intersection is

Example 2: Show that the three lines 3x + 5y + 7 = 0, 2x + 4y + 4 = 0, and 4x – 2y + 18 = 0 are concurrent.

To show the three lines are concurrent, we must show that they all pass through a single point.

 3x + 5y + 7 = 0 3x + 5y = -7 A 2x + 4y + 4 = 0 2x + 4y = -4 B 4x – 2y + 18 = 0 4x – 2y = -18 C

Consider the equations B and C

(2 x B) – C gives

 4x + 8y = -8 – 4x + 2y = 18 10y = 10

10y = 10, so y = 1

Substituting y = 1 in B, we get

2x + (4 x 1)  =  -4

2x  = -4  -4

2x  =  -8

x  =  -4

So the point of intersection of B and C is (-4, 1).

We now substitute this point (-4, 1) in A to check whether the point satisfies the equation 3x + 5y = -7.

LHS – (3 x -4) + (5 x 1) = -12 + 5  =  -7  = RHS

Hence (-4, 1) satisfies equation A, which means the line 1 also passes through (-4, 1). So all three lines pass through (-4, 1), hence they are concurrent lines.