<span lang="EN-US

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