A Geometry Q from GMATPrep S/w

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Digvijay01: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Mon Aug 29, 2011 6:04 am; Thanked: 3 times; Followed by:4 members

A Geometry Q from GMATPrep S/w

by Digvijay01 » Sun Dec 08, 2013 8:37 pm

HI All,
Please help me ans this question. This is a GMATPrep free CAT question. I have attacjed fig. It's not exactly same, but I'll try to explain as much as possible.

In the fig. , 2 lines from points A and B are perpendicular to each other and are of same length. If coordinates of pt A are (-(3)^1/2 , 1), what is the x-coordinate of point B.

1. 3^1/2
2. 1

Note: In actual Qs, these lines are in a semi-circle, unable to draw one here

. Please mention if it might change some result, according to me it should not. Also, I only remember 2 options, 1 is right and 1 is wrong. Will reveal after some discussion.

Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Sun Dec 08, 2013 8:37 pm

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by [email protected] » Sun Dec 08, 2013 10:42 pm

Hi Digvijay01,

In co-ordinate geometry/graphing questions it's important to remember that ANY diagonal line is the hypotenuse of a right triangle (you just have to DRAW in the right triangle).

Draw a line "down" from point A to the X-axis. You now have a right triangle with a base of (root3) and a height of 1. This is a 30/60/90 triangle and you can determine that the line OA has a length of 2. The angle next to the Origin is 30 degrees.

Now, draw a line "down" from point B to the X-axis. You have another right triangle with a hypotenuse of 2.

Since the X-axis is a straight line, it totals 180 degrees and it's made up of 3 angles.

The first = 30 degrees (from the triangle on the left
The second = 90 degrees (from the picture
The third = ????

30 + 90 + X = 180, so the third angle must = 60 degrees.

The triangle on the right is ALSO a 30/60/90. Using the hypotenuse of 2 and the two angles, you can figure out the base and the height of this triangle. Thus, the co-ordinate of point B is (1, root3).

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 am; Thanked: 25 times; Followed by:1 members

by Mathsbuddy » Sun Dec 08, 2013 11:33 pm

The gradient of a normal (perpendicular) to a line with gradient m is equal to -1/m
Using the equation of a line y = mx + c and knowning that c = 0 (through the origin),
then if the first line has co-ordinates (x, mx) at a fixed distance from O, then the perpendicular will have co-ordinates (mx, -x) at the same fixed distance from O.

So the answer is 1.

Last edited by Mathsbuddy on Sun Dec 08, 2013 11:41 pm, edited 1 time in total.

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 am; Thanked: 25 times; Followed by:1 members

by Mathsbuddy » Sun Dec 08, 2013 11:41 pm

Another method:

If a square is rotated about one vertex at the origin, starting off flat against the negative x-axis, then every (x,y) change on the horizontal sides will be duplicated as a (y, -x) change by corresponding points on the vertical sides. In other words, rotation from the x-axis = rotation from the y-axis and the x and y coordinates are interchanged.

Answer = 1

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by GMATGuruNY » Mon Dec 09, 2013 4:47 am

krnverma wrote:

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)Root 2
d)Root 3
e)(Root 2)/2

The sides of a 30-60-90 triangle are proportioned x : xâˆš3 : 2x.
Whenever you see âˆš3 in a geometry question, look for -- or DRAW -- a 30-60-90 triangle.
(-âˆš3, 1) implies a vertical distance of 1 and a horizontal distance of âˆš3.
These distances can serve to form the legs of a 30-60-90 triangle with sides of 1, âˆš3, and 2.
The result is the following figure.

The correct answer is B.

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