Find the equation of the straight line passing through the point of intersection of `2x+y-1=0a d nx+3y-2=0` and making 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)`


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

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

now the given area is `3/8`

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

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