Let’s pick up from where we left last week. We had discussed a coordinate geometry concept using clock faces and had left you with a tough question. Today we will see how you can solve that question using the concepts discussed last week.

You might wonder whether we can expect such a question in actual GMAT. The question we discussed in the last post was an official question and we could solve it easily using this concept. Of course there are many other ways of solving it but this is simplest (or trickiest depending on how you look at it), and it certainly is the fastest, no two ways about it! It is a very logical big-picture approach and people who get Q50-51 often use such methods. The question we will discuss today can also be solved in other ways but we will use the last week’s ‘turning minute hand 90 degrees’ approach.

**Question**: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)

(B) (1, 0)

(C) (1, 1)

(D) (2, 0)

(E) (2, 2)

**Solution**:

First rule of coordinate geometry – draw what you can.

So we make the xy axis and plot the given points, (6, 2) and (0, 6) on it. Let’s say the square is denoted by points ABCD. Say, A is (6, 2) and C is (0, 6). We see that AC is a sloping line. Its two end points are two vertices of the square. We need to find the other two vertices of the square. One of them will lie closest to the origin. The other two vertices will be the end points of the diagonal BD. BD will be perpendicular to AC at the mid-point of AC since a square’s diagonals bisect each other and are perpendicular. So the question is, how do we obtain the end points, B and D? Let’s try to figure out what information we need to draw BD. BD must pass through the mid-point of AC.

How will we obtain the mid-point of AC? By averaging x and y co-ordinates of the points A and C:

x coordinate of mid-point is (0 + 6)/2 = 3

y coordinate of mid-point is (6 + 2)/2 = 4

So BD must pass through (3, 4). When AC turns by 90 degrees, with point (3, 4) as the axis, we get the diagonal BD. So how do the coordinates of AC change when it turns by 90 degrees? Go back to last week’s post and look at the clock face again.

Think of a horizontal line PQ passing through (3, 4).

P coordinate will be given by (0, 4) and Q coordinate will be given by (6, 4) since length of P to mid point is 3 and length of mid-point to Q will also be 3. P shifts up by 2 units to give the point A and Q shifts down by 2 units to give the point C.

Now rotate PQ by 90 degrees and you get RS. We know the coordinates of a line perpendicular to PQ. R will be (3, 1) and S will be (3, 7). This is because R and S will have the same x coordinates as the mid-point (3, 4) and S is 3 units above the mid-point and R is 3 units below the mid-point. We are assuming that you can intuitively see these values on the graph. If not, it may be too soon to spend time on this post.

Now, can we obtain the diagonal BD using RS as reference? If you move S two units to the right, you will get point B (just like A was obtained by moving P two units up) and if you move R two units to the left, you will get point D. Notice that we are using PQ and RS as reference lines. It is easy to calculate vertical/horizontal distances. So B will be given by (5, 7) and D will be given by (1, 1).

The closest co-ordinate to (0, 0) is (1, 1).

**Answer (C).**

Take a few minutes to review the logic discussed here. The ability to ‘see’ such symmetry makes GMAT Quant very simple.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Hi Krishna,

Great fan of your work on this blog as well as on Gmat Club. I wish my Veritas Prep instructor were that clear with concepts and passed on this knowledge to us.

Now to the question – can you please draw a circle on the square diagram so we can get a better understanding. I guess P is sitting at 9:00 clock and B is sitting at 1:00 clock. Please correct me if I’m wrong.

Also I noticed that we went from PQ to RS and then to BD and it became quite easy in the end to envisage the closest vertex of the square. But can you draw a square using AC without the help of PQ and RS, but of course using the same clock concept?

Thanking you.

Regards,

Vaish.

If you note, the same clock concept uses PQ and RS. Notice the colorful lines on the clock face in the previous post.

You are correct that P is sitting at 9 but you cannot define the position of B from that. The position of B depends on the position of A which could be anything as per the given co-ordinates. Notice that here the angle between PM (M being the mid point) and AM is not 30 degrees (sin 30 is 1/2 which is not the case here). So A is not at 10 on the clock face and hence B will not be at 1 on the clock face. The angle between PM and AM could be anything, it doesn’t matter to us. It is easy for us to see how PQ moves to give SR. We use that as a reference to move AC to give BD.

Thank you Karishma for this post. I just came across this question or a similar one and seriously thanked you for this post and explanation.

Because for the life of me, I could not comprehend the explanation given by the other source – pure algebra!

Please keep up the good work and enlighten us with the conceptual ways of solving problems such as this.