Two sides of a square lie along the lines 2y = 20 - 3x and 3x+ 2y = 48 Find the area of the square

Answer 1

the first line contains the point (0,10)
The second line has equation 3x+2y-48 = 0
So the distance between the lines is
|3*0 + 2*10 - 48|/√(3^2+2^2) = 28/√13
That is the side of the square, so its area is 28^2/13 = 784/13

Answer 2

2y = 20 - 3x and 3x+ 2y = 48
y = -1.5 x + 10
and
y = - 1.5 x + 24
those lines are parallel, so the distance between them is the length of a side of your square
the slope of a line between them and perpendicular to them is
-1/-1.5 = 1/1.5
so form of our desired perpendicular is y = x/1.5 + b
a point on first line is (0,10)
so our line can be:
10 = 0 + b or b = 10
y = x/1.5 + 10
where does that hit line two?
x/1.5 + 10 = -1.5 x + 24
14 = x/1.5 + 1.5 x
21 = x + 2.25 x = 3.25 x
x = 6.46
so find y
y = - 1.5 x + 24 = -1.5(6.46) + 24 = 14.3
so our side starts at (0 , 10) and goes to (6.46 , 14.3)
length of side squared = Area = 6.46^2 + 4.3^2

Answer 3

2y = 20 - 3x ... y = -3/2 x + 10

3x+ 2y = 48 ... y = -3/2 x + 16

the slope of the distance between the lines is ... -[1 / (-3/2)] = 2/3

through the lower intercept ... y = 2/3 x + 10

intersection with upper line ... 2/3 x + 10 = -3/2 x + 16
... multiplying by 3/2 ... x + 15 = -9/4 x + 24 ... 13 x = 36 ... x = 36 / 13
... substitute back to find the y-coordinate of the intersection

the distance from the lower y-intercept , (10,0)
... to the intersection with the upper line
... is the length of the side of the square

Answer 4

oops ... the lower y-intercept is (0,10) , not (10,0)