**How to check if 3 lines are concurrent :**

Here we are going to see how to check if given three lines are concurrent.

**Concurrent lines :**

Two or more lines which are passing through same point is known as concurrent lines.

**Point of concurrency :**

The point where all the lines are intersecting is known as point of concurrency.

Here the lines

AB, CD and EF are known as concurrent lines

The point "P" is known as point of concurrency.

**Method 1 :**

(i) Solve any two equations of the straight lines and obtain their point of intersection.

(ii) Plug the co-ordinates of the point of intersection in the third equation.

(iii) Check whether the third equation is satisfied

(iv) If it is satisfied, the point lies on the third line and so the three straight lines are concurrent.

**Method 2 :**

Let the three straight lines be given by

a_{1}x + b_{1}y + c_{1} = 0

a_{2}x + b_{2}y + c_{2} = 0

a_{3}x + b_{3}y + c_{3} = 0

is the condition for the three straight lines to be concurrent.

**Example 1 :**

Show that the straight lines 2x - 3y +4 = 0,9x + 5y = 19 and 2x -7y + 12 = 0 are concurrent. Find the point of concurrency.

**Solution :**

The given equations are

2x - 3y + 4 = 0 ----------(1)

9x + 5y = 19 ----------(2)

2x -7y + 12 = 0 ----------(3)

By solving the 1st and 2nd equation, we get

(1) x 5 => 10x - 15y = -20

(2) x 3 => 27x + 15y = 57

----------------

37x = 37

x = 37/37

x = 1

Apply x = 1 in the 1^{st} equation

2(1) - 3y = -4

2 - 3y = -4

Subtract 2 on both sides

2 - 3y - 2 = -4 - 2

-3y = -6

Divide by -3 on both sides

y = -6/(-3)

y = 2

So the point of intersection of the 1^{st} and 2^{nd} line is (1,2)

Now we have to apply the point (1, 2) in the 3^{rd} equation 2x -7y + 12 = 0

2(1) - 7(2) + 12 = 0

2 -14 +12 = 0

-12 + 12 = 0

0 = 0

From the above statement we understand that the point (1, 2) lies on the lies on the third line.

So the straight lines are concurrent and the point of concurrency is (1, 2).

**Example 2 :**

Show that the straight lines 3x + 4y = 13; 2x − 7y + 1 = 0 and 5x − y = 14 are concurrent.

**Solution :**

If three lines are concurrent,

To prove the given lines are concurrent, we have to convert the given lines in the form ax + by + c = 0

3x + 4y - 13 = 0 -------(1)

2x − 7y + 1 = 0 ------(2)

5x − y - 14 = 0 --------(3)

= 3(98 + 1) - 4(-28 - 5) - 13(-2 + 35)

= 3(99) - 4(-33) - 13(33)

= 297 + 132 - 429

= 429 - 429

= 0

Hence the given lines are concurrent.

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