# <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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

If a + b + c = 0,RD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

Find the equationRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics

<span lang="EN-USRD Sharma - Mathematics