If A_{1}x + B_{1}y + C_{1} = 0 and A_{2}x + B_{2}y + C_{2} = 0 are non-parallel lines then,

**(A _{1}x + B_{1}y + C_{1}) + k(A_{2}x + B_{2}y + C_{2}) = 0**

is the equation of the line through point of intersection of the given two lines.

**Example 1**: Find the equation of the straight line passing through the point of intersection of 2x – 3y + 6 = 0 and x + 3y – 4 = 0, and the point (1,2).

**Solution : Method 1**

Equation of the line through point of intersection is:

(2x – 3y + 6) + k(x + 3y – 4) = 0

This line also passes through (1, 2). Hence putting x = 1 and y = 2 in this equation, we get –

(2 x 1 – 3 x 2 + 6) + k(1 + 3 x 2 – 4) = 0

(2 – 6 + 6) + k(1 + 6 – 4) = 0

2 + 3k = 0, or

k = –.

Now substituting the value of k, we get the equation of the required line is –

(2x – 3y + 6) – (x + 3y – 4) = 0, or

3(2x – 3y + 6) – 2(x + 3y – 4) = 0

6x – 9y + 18 – 2x -6y + 8 = 0

4x – 15y + 26 = 0

**Method 2: **First find the point of intersection of the two equations using simultaneous equations.

2x | – | 3y | + | 6 | = | 0 | + |

x | + | 3y | – | 4 | = | 0 | |

3x | + | 2 | = | 0 |

Or x = –

Putting x = – in x + 3y – 4 = 0, we get

– + 3y – 4 = 0

3y – = 0

y =

Point of intersection is

Now using two point formula for point of intersection and the given point i.e. and (1, 2), we get

y – y_{1} = (x – x_{1})

y – 2 = (x – 1)

= (x – 1)

= (x – 1)

= (x – 1)

y – 2 = (x – 1)

15y – 30 = 4(x – 1)

15y – 30 = 4x – 4

4x – 15y + 30 – 4 = 0

So the required line through point of intersection is 4x – 15y + 26 = 0