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# Line through point of intersection

If A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 are non-parallel lines then,

(A1x + B1y + C1) + k(A2x + B2y + C2) = 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 – y1   = (x – x1)

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