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