Find the equation of the line of slope -3/4 that forms with the coordinate axes a triangle which has the area of 24 square units.

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First the form of the equation would be;

y= mx + b, where m is the slop.

==> y = (-3/4)x + b

when x = 0, then y = (-3/4) 0 + b= b

then the point where the line intersect with the y-axis is (0,b)

when y = 0 ,...

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First the form of the equation would be;

y= mx + b, where m is the slop.

==> y = (-3/4)x + b

when x = 0, then y = (-3/4) 0 + b= b

then the point where the line intersect with the y-axis is (0,b)

when y = 0 , then (-3/4) x +b= 0

==> (3/4) x= b

==> x= (4/3)b

then the point where the line intersect with the x-axis is  (4b/3, 0)

but, the area of the triangle is:

A= 1/2 * 4b/3 * b

24 = 1/2 * 4b/3 *b

24 = 1/2 * 4b^2/3 = 4b^2 /6

==> b^2 = 24*6/4

==> b^2 = 36

==> b = 6, -6

then the eqution of teh slope is:

y = (-3/4) x + 6    and y= (-3/4)x -6

 

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