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Concurrency of Three Lines

We will learn how to find the condition of concurrency of three straight lines.


Definition of Concurrent Lines:

Three or more lines in a plane are said to be concurrent if all of them
pass through the same point.

Concurrent Lines

In the above Fig., since the three lines, m and n pass through the point O, these are called concurrent lines.

Also, the point O is called the point of concurrence.

Three straight lines are said to be concurrent if they passes through a point i.e., they meet at a point. 

Thus, if three lines are concurrent the point of intersection of two lines lies on the third line.

Let the equations of the three concurrent straight lines be

a1 x + b1y + c1  = 0   ……………. (i)

a2 x + b2 y + c2 = 0  ……………. (ii) and

a3 x + b3 y + c3 = 0 ……………. (iii)

Clearly, the point of intersection of the lines (i) and (ii) must be satisfies the third equation.

Suppose the equations (i) and (ii) of two intersecting lines intersect at P(x1, y1). Then (x1, y1) will satisfy both the equations (i) and (ii).

Therefore, a1x1 + b1y1  + c1 = 0 and

a2x1 + b2y1 + c2 = 0               

Solving the above two equations by using the method of cross-multiplication, we get,

x1b1c2b2c1=y1c1a2c2a1=1a1b2a2b1

Therefore, x1  = b1c2b2c1a1b2a2b1 and

y1  = c1a2c2a1a1b2a2b1,  a1b2 - a2b1 ≠ 0

Therefore, the required co-ordinates of the point of intersection of the lines (i) and (ii) are

(b1c2b2c1a1b2a2b1, c1a2c2a1a1b2a2b1), a1b2 - a2b1 ≠ 0

Since the straight lines (i), (ii) and (ii) are concurrent, hence (x1, y1) must satisfy the equation (iii).

Therefore,

a3x1 + b3y1 + c3 = 0

⇒ a3(b1c2b2c1a1b2a2b1) + b3(c1a2c2a1a1b2a2b1) + c3 = 0

 a3(b1c2 - b2c1) + b3(c1a2 - c2a1) + c3(a1b2 - a2b1) = 0

 |a1b1c1a2b2c2a3b3c3|=0

This is the required condition of concurrence of three straight lines.


Solved example using the condition of concurrency of three given straight lines:

Show that the lines 2x - 3y + 5 = 0, 3x + 4y - 7 = 0 and 9x - 5y + 8 =0 are concurrent.

Solution:

We know that if the equations of three straight lines  a1 x + b1y + c1  = 0, a2 x + b2 y + c2 = 0 and a3 x + b3 y + c3 = 0 are concurrent then

|a1b1c1a2b2c2a3b3c3|=0

The given lines are 2x - 3y + 5 = 0, 3x + 4y - 7 = 0 and 9x - 5y + 8 =0

We have

|235347958|

= 2(32 - 35) - (-3)(24 + 63) + 5(-15 - 36)

= 2(-3) + 3(87) + 5(-51)

= - 6 + 261 -255

= 0

Therefore, the given three straight lines are concurrent.

 The Straight Line





11 and 12 Grade Math 

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