find the equation of the straight line passing through the points 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

find the equation of the straight line passing through the points 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


Solution: 

2x+y = 1.....(i)

x+3y = 2......(ii)

on solving (i) and (ii) , we get,

x= 1/5 and

y = 3/5

therefore, are the points of intersection of two given lines.

now, area of triangle in terms of x intercept and y =c intercept.

area of triangle = 3/8

or

here x is coordinate when y = 0 

therefore, 

y = mx+c

0=mx+c

the slope intercept form is 

y=mx+c

the first equation become

and the other equation is 

now, 

thus, other equation is 

y = -12x+3

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