Let’s start with geometry today. It has some very interesting and intuitive concepts. We will discuss one of them today. It’s surprising how a little bit of imagination can go a long way in helping you solve questions. Let’s discuss the concept first. We will look at a question later.

Imagine a clock face. Think of the minute hand on 10. Ignore the hour hand for our discussion today. Say, the length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below (using the green and the red dotted lines). Let’s say the minute hand moves to 1. Can you say something about the lengths of the dotted black and dotted blue lines?

Isn’t it apparent that when the minute hand moved by 90 degrees, the dotted green line became the dotted black line and the dotted red line became the dotted blue line. So can we say that the dotted black line is √3 cm in length and the dotted blue line is 1 cm in length? The same thing will happen when the minute hand goes to 4 and to 7. We don’t think there is much explanation needed here, right? The diagram makes it all clear.

Let’s look at the clock from coordinate geometry perspective. Let’s say the center of the clock is the origin (0, 0). What are the x and y coordinates of the “10’o clock point” i.e. the tip of the minute hand before it moves to 1? Notice that the x coordinate will be -√3 (since the point is in the second quadrant, x coordinate will be negative) and y coordinate will be 1.

What are the x and y coordinates of the “1’o clock point” i.e. the tip of the minute hand after it moves to 1. Notice that the absolute values of x coordinate and y coordinate have switched because the hand has turned 90 degrees. The x coordinate is 1 now and the y coordinate is √3. Since it is the first quadrant, both the coordinates will be positive.

Now, think, what will be the coordinates of the “4’o clock point”, “7’o clock point”? What about the “11’o clock point”, “12’o clock point” etc?

Using this concept, we can solve a very tough GMAT Prep question in a few seconds.

Question 1:

In the figure above, points P and Q lie on the circle with center O. What is the value of s?

(A) 1/2

(B) 1

(C) √2

(D) √3

(E) 1/√2

Solution: You might be tempted to think on the lines of ‘slope of a line’ or ’30-60-90’ triangle (because of the presence of √3) etc. But we should be able to arrive at the answer without using any of those.

Point P is (-√3, 1). O is the center of the circle at (0, 0). When OP is turned 90 degrees to give OQ, the x and y co-ordinates get interchanged. Also both x and y co-ordinates will be positive in the first quadrant. Hence s, the x co-ordinate of Q will be 1 (and y co-ordinate of Q will be √3).

Answer (B)

The question doesn’t seem difficult now (after understanding the concept); actually, it is a 700+ level.

Try another question using the same concept:

Question 2: 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)

*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 Karishma,

Could you please confirm if the options of question 2 are correct. I’m getting (0,2) as the coordinates.

Thanks

H

I think, I jumped the gun a bit quickly. I found my mistake.

Thanks for the nice question.

Thanks