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!*