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Re: My comprehensive Quant Flashcards! [#permalink]

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22 Nov 2017, 11:08

Dear Miguel Thank you very much for the flashcards, they will definitively help me. Please, I have a question for the flascard related to Geometry:coordinate plan, Slide 27: " Consider that the distance between A and B is 4x, and we need a point which is 3x distant from A and x distance from B. Therefore, 3x = 4 --- x = 0.75. The point is located on x-axis -2.75" How Do I get 3x=4? I dont understand that part

This is a nice comprehensive guide - thanks for putting it together.

Two things I noticed:

1) Slide 26 (Geometry - Coordinate Plane): The solution to the equations y = 4x + 10 and 2x + 3y = 26 is x = -2/7 and y = 62/7. The stated solution of x = 4 and y = 6 does not satisfy the equation y = 4x + 10.

2) Slide 27 (Geometry - Coordinate Plane): What are the coordinates for the point on line AB that is 3 times as far from A as from B and that is between points A and B, knowing that A = (-5, 6) and B = (-2, 0)? My solution (below) is different from that in the powerpoint. The methodology in the powerpoint makes an incorrect assumption that the target point, which lays 1/4th of the way between B and A, has an x-coordinate and y-coordinate, each of which are also 1/4th of the way between B and A.

Let point C represent the solution to the question.

Slope and y-intercept of the line connecting A and B: slope: rise/run = (6-0)/(-5--2) = 6 / -3 = -2 y-intercept: y = -2x + b. substituting in x and y coordinates of B gives: 0 = (-2)(-2) + b => b = -4 Therefore equation of the line connecting A and B is y = -2x - 4

The distance between A and B, using pythagorean theorem: Length of side 1 = absolute value of -5-(-2) = 3 Length of side 2 = absolute value of 6 - 0 = 6 Length of hypotenuse = sqrt(9+36) = sqrt(45) = 3 sqrt(5)

Therefore we know that the distance between B and C (i.e., the hypotenuse of the right triangle with vertices at B and C) is: 3 sqrt(5) / 4 Given the slope of the line containing A, B and C is -2, we know that the right triangle with vertices at B and C has legs in a 1:2 ratio.

Let x = the length of the horizontal leg of the right triangle with vertices at B and C. Let y = the length of the vertical leg. We know that y = 2x. Therefore: (x)^2 + (2x)^2 = 3 sqrt(5) / 4 5x^2 = 3 sqrt(5) / 4 x^2 = 3 sqrt(5) / 20 x = sqrt(3 sqrt(5)) / (2 sqrt(5)), which is approximately equal to 1.15829

Therefore the coordinates of point C can be represented in relation to point B: C = (-2 - x, 0 + 2x), in which x is the irrational number which we calculated above.

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