Question 1110393
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It is not specified in the statement of the problem, but I will assume the triangular region is in quadrant I.  There is a second, corresponding solution with the region in quadrant III.<br>
With the gradient of -3/4, we can let A = 4x and B = 3x.<br>
Then the area of the triangle is {{{(1/2)(4x)(3x) = 6x^2}}}.<br>
Since the area is 54,
{{{6x^2 = 54}}}
{{{x^2 = 9}}}
{{{x = 3}}}<br>
Then A is 4x=12 and B is 3x=9.<br>
Now we have the gradient and the y-intercept; the equation of the line is {{{y = (-3/4)x+12}}}<br>
The other solution comes from choosing x = -3 instead of x = 3 when solving x^2=9.  The result is A = -12 and B = -9; the equation of the line is {{{y = (-3/4)x-12}}}.