The vertex of a parallelogram are (1, 0), (3, 0), (1, 1) and (3, 1)

First, quickly draw an xy graph and draw the rectangle.



Look at the answer choices. You should see that B and D are immediately out. Slopes of 2 and 3 (or y = 2x and y = 3x) will miss the rectangle completely. Visually, this should take a second.

Now let's try answer choice A of slope 1/2 (y = .5x). It's the middle of 1/4 and 3/4. If the line has a slope of .5, then it must go through points (1, 0.5) and (2, 1) through the rectangle. Visualize that line and see where it goes through on the rectangle. Clearly, this doesn't divide the rectangle in half like we wanted, so answer choice A is out.

Answer choice E is out as well, since it's greater than 1/2. We need a smaller slope.

So at this point, I would put down answer choice C. But let's verify anyway. Slope 1/4 is the same as y = .25x, and it should be a line going through (2, 0.5). Since this is the center of the rectangle, it makes sense this would be the answer.

C it is!

You also might find helpful the links in my signature.

Here's an alternative way to reach the answer using the equation for a line: \(y-y_1 = m(x-x_1)\)

By visualizing the line you can see that in order to divide the rectangle evenly it needs to intersect the rectangle's right side (i.e. it enters on the left and leaves on the right).

Let t be the y-value of the line at x = 1. Using this definition, the y-value at x=3 must be 1-t, since the height of the rectangle is 1. Note that if it weren't 1-t, the division of the rectangle would not be even.

This gives two points on the line: (1,t) and (3,1-t). So we have two equations for the same line:

(1) \(y-t = m(x-1)\) [using the point (1,t) ]
(2) \(y-(1-t) = m(x-3)\) [for the same line using the point (3, 1-t) ]

Since this is the same line, we can solve for m, the slope. Solving for t in (1) gives \(t = y-m(x-1)\)

Substituting this value of t in (2) gives:

\(y-(1-(y-m(x-1))) = m(x-3)\)
\(y-1+y-m(x-1)=m(x-3)\)
\(2y-1=m(x-3)+m(x-1)\)
\(2y-1=m((x-3)+(x-1))\)
\(\frac{(2y-1)}{((x-3)+(x-1))}=m\)

Since the origin is on the line, plug in x=0, y=0:

\(\frac{-1}{(-3-1)}=m =\frac{1}{4}\)

Note that plugging in x = y = 0 before the algebra would work as well, but I included the algebra because it's more illustrative.

Quickest method:

Attached image is borrowed from farful 's post above.

Added the line BE (the required line that divides the given parallelogram into 2 identical quadrilaterals).

Careful observation reveals that slope of the diagonal AD = (1-0)/(3-1) = 0.5. Thus the slope of line BE should be < slope of AD. Only option C is < 0.5 and is thus the correct answer.

Hi,

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your answer does not seem to be correct..
ans should be C. 1/4..
You can draw the quadrilateral and evevn the coord suggest that it is a rectangle...
the height of rectangle is 1 and length is 2...
1)this line passing through origin cuts the left side at say y... coord = (1,y)..
SLOPE with origin (0,0) = \(\frac{y-0}{1-0} = y\)
2)for the two quad to be similar, the line should cut the right side at y units below the upper vertex.. so coord = ( 3, 1-y)...
SLOPE = \(\frac{1-y-0}{3-0} = \frac{1-y}{3}..\)

but bothslopes should be SAME..
\(y = \frac{1-y}{3} .............3y = 1-y..........y = \frac{1}{4}....\)
so slope = y = 1/4
so sl

I tried and i am getting the answer as 1/4 (option C). Are you sure about the OA??

(Please refer the rough figure attached)
As we can see, the line L passes from below the point (3,1). Now, had the line passed through (3,1) the slope would have been 1/3. (line passing through (0,0) and (3,1)). Since our line L is passing from below (3,1) the slope has to be less than 1/3.

The only option less than 1/3 is option C 1/4.

Option C

A good understanding of what slope means helps you solve this question in seconds. Draw out the parallelogram and you see that it is a rectangle with base at the x axis between 1 and 3. You want to split the rectangle into identical quads using line L passing through centre. So it will look like the diagram drawn by Engr2012 here: the-vertex-of-a-parallelogram-are-1-0-3-0-1-1-and-218118.html#p1561948

Now use the concept of slope, say the slope of line is n - the y co-ordinate increases by n every time the x co-ordinate increases by 1.
When the line moves from 0 to 1 x co-ordinate, its y co-ordinate increases by "n" (so AB is n). When it moves from x co-ordinate 1 to 3, y co-ordinate will increase by 2n. To make the quads identical, DE should be n too.
Hence the side of length 1 is split into n + 2n + n.
This means n = 0.25 = 1/4

Answer (C)

A discussion on slope: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/04 ... line-gmat/

The line should not pass through the farthest corner if that happens the quadilateral will not be equally divided therefore it should be greater then 1/3

2 and 3 goes beyong the recrangle therefore out

1/2 will not divide into half and the only possibilityleft is 1/4

THerefore IMO C

Another way to look at this problem is from the viewpoint of the line that will divide the figure into two equal parts.
Here, we have a rectangle, and a diagonal divides the rectangle into two equal parts. The diagonal passes through the mid-point of the two alternate vertices. The mid-point of the diagonal in this case is, say Point X, (2, 1/2).

Now, logically, any line dividing a rectangle into two equal parts must pivot or pass from the point X.
The slope of the line passing through the point X and Origin is 1/4, which is given in option C.

Check: https://youtu.be/Ys_tKG7dQ6o