Linear Equations: Point of Intersection of Lines

The point of intersection of two or more lines is a point which lies on all the given lies. It means the equations of all the given lines must be satisfied by the intersection point. This point of intersection of lines is called the “point of concurrency”.  Finding this point of concurrency of two lines from given set of lines is used to determine whether the other lines are concurrent with these two lines.

Point of intersection of two lines:

Let two lines a₁x+b₁y+c₁ =0 and a₂x + b₂y + c₂=0 represent two intersecting lines.

Let the intersecting point of these two lines be (x₁,y₁).

Then it must lie on both the lines.

a₁x₁+b₁y₁+c₁=0

a₂x₁+b₂y₁+c₂=0

Solving these two equation, we get

(x₁,y₁)=([b₁c₂-b₂c₁]/[a₁b₂-a₂b₁], [c₁a₂-c₂a₁]/[a₁b₂-a₂b₁])

Example:

Find the point of intersection of lines 2x+3y-8=0 and x-y +1=0.

Solution:

Here a₁=2, b₁=3, c₁=-8 and a₂=1, b₂=-1, c₂=1,

Pint of intersection (x,y) = ([b₁c₂-b₂c₁]/[a₁b₂-a₂b₁], [c₁a₂-c₂a₁]/[a₁b₂-a₂b₁])

=([3-8]/[-2-3], [-8-2]/ [-2-3])=(-5/-5, -10/-5)=(1,2).

Condition for point of concurrency of 3 lines:

Let the 3 lines a₁x+b₁y+c₁ =0, a₂x + b₂y + c₂=0 and a₃x + b₃y + c₃=0 be concurrent. Then the point of intersection of first two lines must lie on the third line.

i.e. ([b₁c₂-b₂c₁]/[a₁b₂-a₂b₁], [c₁a₂-c₂a₁]/[a₁b₂-a₂b₁]) must line on the line a₃x + b₃y + c₃=0.

So, a₃ [b₁c₂-b₂c₁]/[a₁b₂-a₂b₁]  + b₃ [c₁a₂-c₂a₁]/[a₁b₂-a₂b₁]  + c₃ = 0

Then we get

a₃ [b₁c₂-b₂c₁] + b₃ [c₁a₂-c₂a₁] + c₃ [a₁b₂-a₂b₁] = 0.

Example:

Determine whether the lines 2x+3y-8=0 , x -y +1=0 and 3x + y-5=0.

Solution:

a₁=2, b₁=3, c₁=-8 , a₂=1, b₂=-1, c₂=1 and a₃=3, b₃=1, c₃=-5.

Consider a₃ [b₁c₂-b₂c₁] + b₃ [c₁a₂-c₂a₁] + c₃ [a₁b₂-a₂b₁]

= 3 (3 – 8) + 1 (-8-2) + (-5) (-2-3)

= -15 -10 + 25

=0

Since the condition for concurrency is satisfied, the given lines are concurrent.

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