I would say it hit the square at (a, 0) and at (b , 1)
Area of trapezoid on left = (a+b)/2(1) = (a+b)/2
Area of trapezoid on right = [(1-b)+(1-a)]/2
= (2 - a - b)/2
if those areas are equal then
a + b = 2 - a - b
or as we could see from the sketch
a + b = 1
Now the slope of 6
b = a + (1/6)
a + a + 1/6 = 1
2 a = 5/6
a = 5/12
b = 5/12 + 1/6 = 7/12
NOW you have a problem I am sure you can do
find c in y = 6 x + c
It goes through (5/12 , 0)
0 = 5/2 + c
c = -5/2
y = 6 x - 5/2
for y axis intercept, x = 0
y = -5/2
You know the line is y=6(x-a) so it intercepts the x-axis at (0,a), and the line y=1 at x=(1+6a)/6 = 1/6 + a
So, the left half of the figure is a trapezoid with height 1 and bases a and (1/6 + a)
So, if that has area 1/2,
1(a+a+1/6)/2 = 1/2
a = 5/12
so, y = 6(x-5/12)
y = 6x - 5/2
Or, consider the point C=(1/2,1/2), the center of the square. A vertical line through C would cut the square in half.
Since we want a line of slope 6, so y increases 6 for every increase in 1 by x. If the line passes through C, then if it leans to the right 1/12 at y=1 and to the left 1/12 at y=0, it will still cut the square in half. So, we have the line through (5/12,0) and (7/12,1).
y = 6x - 5/2
THANKS A LOT!
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