Question

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

Answer 1

The point of intersection of     2 x + y = 1  ---  (1)        and       x + 3 y = 2    -- (2)    is  found by   (1) * 3 -  (2) .
       6x + 3 y - x - 3y = 3 - 2
       =>  x = 1/5    and     y = 3/5  using (1)
Point of intersection is A (1/5,  3/5)

See the diagram attached.

Equation of a line with an x intercept of a and y intercept of b is:
        x / a + y / b  = 1          ---  (3)
 As the point A lies on the above straight line,
        1/(5a)  + 3/(5b) = 1      --- (4)

Area of the triangle formed is = 1/2 * | a b |  = 3/8    --- (5)
             So  either  a* b  = 3/4      =>    b = 3/(4a)    --- (6)
                        or    a * b  = - 3/4    =>  b = -3/(4a)    --- (7)

We use equation (4) and  (6) together:
             1/(5a) + 3/5 * 4a/3 = 1
             1/a + 4a = 5        =>  4 a² - 5 a + 1 = 0
           =>  (4a - 1) (a -1) = 0  =>  a = 1 or  1/4
             =>  b = 3/(4a) = 3/4  or  3

 So the equations of two possible lines are:
        x + 4/3 y = 1    =>   3x + 4y = 3           --  line A in the diagram.
         4 x + y/3 = 1    =>  12 x + y = 3          -- line B in the diagram.
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Now, we use the equations (4) and (7) together:
         1/(5a) - 3/5 * 4a/3 = 1
          1/a - 4a = 1
           4 a² + a - 1 = 0      =>  a  = [ -1 +- √(1+16) ] /8
                   a = (√17 - 1)/8      or      -(√17 + 1)/8
       =>      b = -6/(√17-1)        or      6/(√17+1)

Hence the equations of these two straight lines are :
   8 x /(√17-1)  - (√17-1) y/6 = 1  =>  24 x - (9 - √17) y = 3(√17 -1)          -- Line C
  - 8x/(√17 + 1) + (√17+1) y/6 = 1  => - 24 x + (9+√17) y = 3(√17+1)      -- Line D.

So there are 4 lines.  Two in the 1st quadrant, one in the second and one in the fourth quadrant.