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

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