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April 16, 2014

April 16, 2014

Posted by **Knights** on Saturday, March 9, 2013 at 5:28pm.

I tried putting one point where the line intersects the square as (y,1), and the other as (x,0), and the y intercept as (0,a). Then, I tried using the slope to figure out the variables. The only problem is that I dont have enough equations. Help?

- Please help analytic geometry -
**Damon**, Saturday, March 9, 2013 at 5:52pmI 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)

so

a + a + 1/6 = 1

2 a = 5/6

a = 5/12

then

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

so

c = -5/2

y = 6 x - 5/2

for y axis intercept, x = 0

y = -5/2

- Please help analytic geometry -
**Steve**, Saturday, March 9, 2013 at 9:46pmYou 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

- Please help analytic geometry -
**Knights**, Saturday, March 9, 2013 at 9:49pmTHANKS A LOT!

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