Question 1094438
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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 {{{-1/5}}} and has, therefore,
        the equation y = {{{(-1/5)*x}}} + 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,{{{c/5}}}).


It means, that your right-angled triangle has the legs of  {{{abs(c)/5}}}  and  |c|  units long.

Then its area is  {{{(1/2)*(abs(c)/5)*abs(c)}}} = {{{c^2/10}}} square units.


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

{{{c^2/10}}} = 5,

which implies  {{{c^2}}} = 50  and then  c = {{{sqrt(50)}}}.


It means that your final equation under the question is

5y + x = {{{sqrt(50)}}}.
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Solved.