Find the equation of the straight line which passes through the point of intersection of the lines `3x-y=5\ a n d\ x+3y=1\ ` and makes with the coordinate axes a triangle of area` 3/8` sq. units.

Explanation

The equation of straight line passing through the point of intersection of` 2x+y−1=0` and` x+3y−2=0` is

`2x+y−1+lambda(x+3y−2)=0`

`(2+lambda)x+(1+3lambda)y−1−2lambda=0.....`(i)

`x/[(1+2lambda)/(2+lambda)x]+y/[(1+2lambda)/(1+3lambda)]​=1`

So the point of intersection of this line with coordinate axes are

`((1+2lambda)/(2+lambda)​,0) and (0,(1+2lambda)/(1+3lambda)​)`

It is given that the required line makes an area of 3/8​ sq. units with the coordinate axes.

`1/2​(1+2lambda​)/(2+lambda)×(1+2lambda)/(1+2lambda)​=3/8`

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