A certain computer program randomly generates equation of line is form of y=mx+b.If point p is a point on a line generated by this prog, what is probability that line does not pass through ABCD.

A. 3/4
B. 3/5
C. 1/2
D. 2/5
E. 1/4

I am responding to a pm from himanshuhpr.

First of all, I am intrigued by the solution of the very intelligent VeritasPrepKarishma, for whom I have considerable respect. Nevertheless, I beg to differ.

I believe there are some significant flaws in the question. First of all, the grammar is embarrassingly atrocious, but we'll let that pass. More to the point === we are told that the program "randomly" generates lines --- how exactly is this process randomized? For example, the line goes through P --- Are all slopes equally likely? Or are all angles equally likely? If we select P and a y-intercept, are all y-intercept equally likely? Are we randomly selecting one other point in the x-y plane, and all of those points are equally likely? Those are four different ways of specifying the probabilistic process, and I believe at least some of them give different numerical answer to the question. As soon as the thing you are specifying (here, a line) can be specified in a variety of ways, its not enough to say "random" --- you have to specify exactly how the selection process proceeds.

Because the question does not specify the details of the probabilistic selection process, I believe it is a fundamentally flawed question.

Does that make sense?

Mike
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Dear himanshuhpr,

I understand your desire to understand. As I explained in my pm response, I think this question is too flawed to admitted of an unambiguously good solution.

Instead, I created a similar question, and I believe in this question the entirely selection process of the line is very tightly defined, making the question unambiguous. Here's where it's posted ......

hard-probability-problem-lines-in-the-x-y-plane-142846.html

After some folks wrestle with it, I will post a solution.

I hope that's helpful.

Mike
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Interesting... while I do agree the question is poorly worded. I was able to come to (C) as the answer, but I had to recall my SIN/COS/TAN angles...which seems overkill for the GMAT. Even for a 700+ question...

Step 1:
The graph shows point P(6 units, 0 units), using either points C(4 units, 2 units) or D (8 units, 2 units) you can derive a triangle with two sides of length 2 units, and a hypotenuse we don't need to know.

Remembering SOH/CAH/TOA -> Use TOA. Tan of Angle (Degrees) = Opposite Length / Adj. Length = 2 / 2 = 1. Tan of 45 Degrees = 1.

Therefore, the sector bounded by the two linear equations is 180 - (45 x 2) = 90 degrees.

Step 2:
Realize that it will not simply be 90 / 360 (= 1 / 4) as this will include linear equations that cross ABCD.

Using the two bounded lines, visualize a North/South/East/West quadrant and realize that North/South sectors are covered by the same line equations, East/West sectors are covered by other line equations.

Probability of Lines HITTING ABCD = (Sector covered by North and South quadrant) / 360 = (90 x 2) / 360 = 1 / 2.

Probability of Lines NOT HITTING ABCD = 1 - (1 / 2) = 1 / 2.

Answer (C).

I took an approach similar to jcaine:

Looking at the figure, you can determine that the triangle formed by DPC has a 90 degree angle at P.

This can be assumed because triangle formed from D, (x = 4) , P is a 45-45-90 triangle. This is certain because the letters fall on distinct points within the graph. A 45-45-90 triangle is also formed by C, (x=8), P. The sides of each triangle formed here are 2-2-x SQRT 2 (45-45-90 triangle)

Using the property of a straight line = 180 degrees, you can deduct that triangle DPC has a 90 degree angle at P and any intersecting line would fall within 90 degrees out of 180.

Double that to get 180 / 360 degrees = 1/2.

Think of it as a circle and the "range" (degree) that the line can and can not pass through.

I hope this explanation helped.

Now I see what the Q meant to test here...Super horrible grammar..
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Point D coordinates (4, 2)
Point C coordinates (8, 2)
So if you extend AD and BC to touch the x axis, you get a rectangle (shown in my figure). Split it into two equal rectangles by dropping a perpendicular from P to DC. Now draw the diagonals of both rectangles as shown. Note that the diagonal of a rectangle divides the rectangle into two equal areas. Since the two rectangles were of equal areas, all four areas will be equal. Basically , you divide the entire region around point P into 4 equal areas - in 2 of them, the line will pass through the square, in 2, it will not.
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let O be the point on CD at right angle from P
apply pyth. thm. on triangles POC and POD.. Now we can notice that PC = PD
hence PCD is a isosceles triangle
therefore Angle CPD = 90
So if 0<= Angle CPD <=90 then line will cross the rectangle
So favorable outcomes are 90 and possible outcomes are 180

so probability will be 90/180 = 1/2
probability that line will not cross through rectangle will be 1 - 1/2 = 1/2

Hi Karishma,

How did you arrive at the 4 lines ( red and green)? Please explain.

Hi

I think y = mx+b is just given to state general equation of a line. (meaning this line has a slope of 'm' and a y intercept of 'b')
Using this equation is not required here, as you can see from the previous solutions.