Find the equation of the straight line passing through the point of intersection of 2x + y – 1 = 0 and x + 3y – 2 = 0 and making with the coordinate axes a triangle of area 3/8 sq. units.
Given:
2x + y – 1 = 0 and x + 3y – 2 = 0
To find:
The equation of the straight line passing through the point of intersection of 2x + y – 1 = 0 and x + 3y – 2 = 0 and making with the coordinate axes a triangle of area 3/8 sq. units.
Explanation:
The equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 is given below:
2x + y − 1 + λ (x + 3y − 2) = 0
⇒ (2 + λ)x + (1 + 3λ)y − 1 − 2λ = 0
⇒ 
So, the points of intersection of this line with the coordinate axes are 
and 
It is given that the required line makes an area of 
square units with the coordinate axes.
⇒ 3 |3λ2 + 7λ + 2| = 4 |4λ2 + 4λ + 1|
⇒ 9λ2 + 21λ + 6 = 16λ2 + 16λ + 4
⇒ 7λ2 – 5λ – 2 = 0
⇒ λ = 1, 
Hence, the equations of the required lines are
3x + 4y – 1 – 2 = 0 and 
⇒ 3x + 4y – 3 = 0 and 12x + y – 3 = 0