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

(ii) Plug the coordinates 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.

**Example 1 :**

Prove that the straight lines

2x+y = 1, 2x+3y = 3 and 3 x + 2 y = 2

are concurrent. Find the point of concurrency.

**Solution :**

The given equations are

2x + y = 1 -----(1)

2x + 3y = 3 -----(2)

3x + 2y = 2 -----(3)

Multiply the 1^{st} equation by 3 and subtract the 2^{nd} equation from 1^{st} equation.

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

6x+3y-2x-3y = 3 - 3

4x = 0

Divide 4 on both sides

4x/4 = 0/4

x = 0

Apply x = 0 in the 1st equation, we get

2(0) + y = 1

y = 1

Now let us apply the point (0, 1) in the third equation

(3)===> 3 x + 2 y = 2

Here x = 0 and y = 1

3 (0) + 2 (1) = 2

0 + 2 = 2

2 = 2

Since the point (0, 1) satisfies the 3^{rd} equation, we may decide that the point(0, 1) lies on the 3^{rd} line.

Hence the given lines are concurrent and the point of concurrency is (0, 1).

**Example 2 :**

Prove that the straight lines

3 x + y = -2, 2 x - y = -3, x + 4 y = 3

are concurrent. Find the point of concurrency.

**Solution :**

The given equations are

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

2x - y = -3 -----(2)

x + 4y = 3 -----(3)

Adding 1^{st} and 2^{nd} equation we get,

3x + y + 2x - y = -3 + (-2)

5x = -5

Divide by 5 on both sides

5x/5 = -5/5

x = -1

Apply x = -1 in the 1st equation, we get

3(-1) + y = -2

-3 + y = -2

Add 3 on both sides we get,

-3 + 2 + y = -2 + 3

y = 1

Now let us apply the point (-1, 1) in the third equation

(3)===> x + 4 y = 3

Here x = -1 and y = 1

-1 + 4 (1) = 3

-1 + 4 = 3

3 = 3

Since the point (-1, 1) satisfies the 3^{rd} equation, we may decide that the point(-1, 1) lies on the 3^{rd} line.

Hence the given lines are concurrent and the point of concurrency is (0, 1).

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