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. Hint - there are two eq

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 of the same form 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 you finally have two equations to answer your question

5y + x =   and  5y + x = .