SOLUTION: Find the equation of a straight line that is perpendicular to 5x–y=1 and is such that the area of the triangle formed by the x- and y-axes is equal to 5.

Straight line perpendicular to  5x-y = 1  has an equation

5y + x = c,   (1)

where c is some (arbitrary) constant.


        Our original line has the slope 5; so, the perpendicular line has the slope  and has, therefore,
        the equation y =  + c,   which is the same as (1).


So, all you need to do is to determine the constant "c" in equation (1).


For it, notice that straight line  (1)  has x-intercept  (c,0)  and y-intercept  (0,).


It means, that your right-angled triangle has the legs of    and  |c|  units long.

Then its area is   =  square units.


You need to have this area equal to 5 square units. It gives you an equation

 = 5,

which implies   = 50  and then  c = .


It means that your final equation under the question is

5y + x = .