The vertices of a rectangle in the standard (x,y) coordinate place are

It is not possible to have the point (2,2) on the diagonal. Had it been on the diagonals, the slope of this line would be : \(\frac{4-0}{7-0} = \frac{2-0}{2-0}\) which is obviously not the case as these 2 values are different.

First I assumed the line passes through the origin and is a diagonal of the rectangle making the slope 1. But then I realized that the slope can't be '1' because only a square would have a slope of 1. Since this is a rectangle, its slope has to be something else.

This is a good problem where the grid lines on the worksheet come in handy. Just need to make sure to draw the sketch to scale.

A diagonal is not the only line that divides the area into two equal halves. Nor is the horizontal line. A vertical line could also divide a rectangle into two equal facts. In fact, if there is only one point given within a rectangle, we can draw a line that would not be horizontal, vertical or diagonal, that would divide a rectangle into two trapeziums of equal area. However, we need to identify which of the following points are on a line that divides the rectangle into two equal halves. A figure approach, as demonstrated by bunuel, helps.

However, if you wish to check through an extended version, find the equation of the line that would make the diagonal of the rectangle. You'd find that the point (2,2). does not lie on the diagonal. Also, it would not lie on a line that is vertical and that divides the rectangle into two equal halves. Another way of looking at the question is that the breadth of the rectangle is 4 and the length is 7. So a horizontal line with y co-ordinate 2 will for sure divide the rectangle into two equal halves. Also, a vertical line with x co-ordinate 3.5 would do the same job.

Hope this helps.

Why did you not check to see if it is the diagonal of the rectangle?
Is it not possible for the diagonal to split it into 2 equal shapes?

Yes this is what I thought.
I just didn't understand if it was a given that we need to check it, or if there was
another way of knowing without making this equation and checking.

Query
Hi Bunuel,

"In order the line to divide the rectangle into two equal parts it must be horizontal. The slope of any horizontal line is zero" Why is this always true?

Why cant a line that pass through (2,2) and make 45 degrees(slope 1) with X axis and that also splits the rectangle into two quadrilaterals be assumed ?

Line passes through 2,2
Equation of line=y-y1=m(x-x1)
So y-2=m(x-2)

From the options I initially choose m=1
But it will intersect the line segment joining 0,4 and 7,4 at x=4,y,4
Two areas will be unequal

next chose m=0,
It will be a horizontal line

So,y=2 is the equation of the line and it will intersect line joining 7,4 and 7,0 at 7,2
Now we have two rectangles and both of them have areas=14.

So m=0 is the answer.

I am unable to understand why cant it be a diagonal and why horizontal line. The rectangle has 1 pair of sides equal , the diagonal from origin passing thru 2,2 will divide into 2 parts that is equal. Please tell me why we are considering horizontal line and not the diagonal.

Regards
Megha

I was initially confused that for splitting the triangle into 2 equal areas, diagonal might be required. But then, when the graph is plotted, the line that goes through (2,2) will go straight and have coordinates (0,2) & (7,2). The area formed is half of the rectangle and the slope would be 0. [Distance between two points (2,2) - (0,2)].

In the above graph as shown by Bunuel

the line through (2,2) is parallel to both axis of rectangle rectangle

We know the product of slopes for parallel lines [(m1* m2)=0]

so the slope of the line through (2,2) is 0 (since slopes of other lines can not be zero)

I hope it's clear

Note: You can think of a 45-degree solution, but answer not in option list.

M=0 (horizental line) can be obvious once we draw the rectangle and point (2,2).

In case we didn't have M=0 as an optional answer we should have looked for line that goes through (2,2) and devide the rectangle to two equal-sized trapezoids.

I wonder, what is the 'safest' and quickest way to find this line?

Thanks for the answers

If you drag the diagram out on your scratch pad accurately then this should be very straight forward as the only line that could possibly divide this rectangle into two equal areas is a horizontal line through 2,2; thus, the slope must be 0.

Thatll make it a square..

Recall that a horizontal slope = 0; a vertical slope = undefined.

Given the coordinates (2,2), the only way to split the rectangle evenly is to have a horizontal line going through the rectange.

Therefore we can conclude the slope = 0.

This is how I got it, once I realized the line was through point 2,2 I knew it couldn't be vertical or through the vertices because both parts weren't even. As Bunuel stated, if the point was (3.5, 2), then the line could be in a few different directions with different slopes because that's dead middle of the rectangle.