Of the following, which are the coordinates of a point that is located

Hi chetan2u, amanvermagmat

From the equation Slope m = tanx = 2.... From Option A & C tanx = 1.25 and 2.4.... more the value of m, more the angle x will incline towards Y-axis.. less the value of m.. more the angle will incline towards x-axis.. So option A.

Can you please provide your feedback on this or am I missing something??

The line passes through two points: (3, 6), (0, 0)

Equation of this line is y = 2x (since slope is 2 and y intercept is 0)
So y - 2x = 0

Consider any point on the left side of the line e.g. (-1, 0). If we plug it in y - 2x, we get 2 which is positive.
So our shaded region is represented by y - 2x > 0 and y < 0.

So options (B) and (D) are straight away out (since y is positive in them).
Option (A) satisfies y - 2x > 0 and hence is the answer.

For more on this approach, check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/1 ... he-graphs/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/1 ... s-part-ii/

Hi Rahul

Slope of a line is calculated from two points on the line, not from a single point. I think you are calculating slope based on a single point only. From your query I assume that from option A, you have calculated slope as -5/-4 = 1.25, this is not how we find slope because slope of a single point is meaningless. Slope is always about join of two points (since this join will form a line), and with the slope we can get some idea about how the join of these two points will tilt and look like. Slope of a line passing through points (x1, y1) and (x2, y2) is = (y2-y1)/x2-x1) Or (y1-y2)/(x1-x2)

The line which is given in the diagram, line m - we know this line passes through the origin (which is 0,0) and another point (3,6). So its slope is (6-0)/(3-0) = 6/3 = 2.
Now our required point has to be in the shaded area, and the entire shaded area lies in 3rd quadrant, where both x and y coordinates are negative. So out of the given options we have only two to consider A and C options, where the points given are (-4, -5) and (-5, -12) respectively.

If we look at option C. Now imagine a line which connects the point (3,6) to (-5, -12). The slope of this line will be = (-12-6)/(-5-3) = 18/8 > 2. Since the slope of line formed like this is positive but greater than that of line m, this line will tilt further towards y-axis than the given line m (when comparing two positive slopes, the higher the slope the steeper the line). Because of this tilt, in the first quadrant this line will lie to the left of line m, and in the third quadrant this line will lie to the right of line m. Which automatically leaves the shaded area (doesn't include it). Hence option C point (-5, -12) cannot lie in the shaded area.

Now lets look at option A, if we join the point (3,6) with point (-4, -5) then the slope of line so formed will be = (-5-6)/(-4-3) = 11/7 < 2.
Since slope of this line is positive but less than that of line m, this line will tilt away from y-axis than line m (since slope of m is greater, it will be steeper than this line). So in the first quadrant this line will lie to the right of line m, and in the third quadrant this line will lie to the left of line m. Because of lying to the left of line m, this will definitely intersect with the shaded area there. Hence we can conclude that this point (-4, -5) will lie in the shaded area.

Tried to get it without paper work. Shaded region is in 3rd quadrant so (-x, -y). Leaves us with options A & C.

Slope of the line is 2. So lines from origin (0, 0) to any point (-x, -y) on the shaded region should have slope less than 2.

Only option A satisfies the above condition and hence it is the right choice.

Just restarted my GMAT life. KUDOs please

My approach:
Since the shaded area is in 3rd quadrant, both the co-ordinates shall be -ve.
Eliminate B, D & E.
Left with A & C
Line M has an ascending slope.
The unshaded area in the 3rd quadrant will increase more and more as we go down on y axis i.e far from the origin on y axis.
Co-ordinate -12 (in option C) on y-axis is the farthest than co-ordinate -5 (in option A).
Thus Option A.
Bunuel.. Is my logic correct?

CAN BE DONE WITHOUT PAPER .SHADED REGION IN 3 QUAD.THEREFORE BOTH X AND Y INTERCEPT SHOULD BE NEGATIVE.THIS BRINGS DOWN TO 2 CHOICES EITHER A OR C.
slope =2
shaded region will have slope less then 2 ,only A satisfy this condition .hence ans is A

Hi VeritasKarishma,

I totally understand this concept. However, my doubt is that instead of taking y-2x, if we take 2x-y and substitute -4,-5, we see that 2x-y becomes < 0. Can you please explain where am I mistaking in my concept here? y=2x can be both y-2x=0 and 2x-y=0.

Note that y - 2x = 0 or 2x - y = 0 is the equation of the line (of course, both are same). We need to find how to represent the shaded region bounded by this line and x axis.

(-1, 0), which lies in the shaded region, makes 2x - y < 0 so the shaded region is represented by 2x - y < 0 and y < 0 (since it is below the x axis).

Now check which of the given five options satisfies these conditions. Only (-4, -5) does.

Thanks a lot VeritasKarishma,

Sorry for the separate mail thread. I was not allowed to continue in the same thread.

So what I understand is that the shaded region will be represented either as 2x-y<0 and y<0 or y-2x<0 and y<0. Can you please explain why is the inequality sign getting reversed w.r.t the equation? The concept that i knew was that the region above the graph is positive and that below the graph is negative. However can you please what is the dependence between the variables and the inequality sign?

Are you saying that we could test the inequality with a simple sure shot point in the region and decide on the inequality sign w.r.t the expression used?

A line is represented by y = mx + c. This is an equation. This divides the co-ordinate plane into two parts. One will be y < mx + c and the other will be y > mx + c. Usually, we plug in (0, 0) into the line and see which inequality holds. Now the region in which (0, 0) lies is the region of that inequality.

Note here that our shaded region is represented by either 2x - y < 0 or by y - 2x > 0. In both cases, it is just 2x < y which is the same as y > 2x. It has to be the same because we are talking about the same region.
So our line is y = 2x, and our shaded region is y > 2x.

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