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The equation of line m is y=x/2+1. What is the distance from the xint
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12 May 2015, 13:01
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The equation of line m is y=x/2+1. What is the distance from the xintercept of m to the intersection of x=4 and m? A. √(13) B. 5 C. 6 D. 3√5 E. 7
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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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12 May 2015, 22:18
Hi reto, To answer this question, we'll first need to figure out the two coordinates involved.... Given the line Y = (X/2) + 1 The Xintercept will be the point at which Y = 0 0 = (X/2) + 1 1 = X/2 X = 2 So the first coordinate is (2, 0) The second coordinate is where X = 4.... Y = (4/2) + 1 Y = 3 So the second coordinate is (4, 3) The diagonal line segment that exists between these two points is the hypotenuse of a right triangle. That triangle has a base of 6 and a height of 3. So, with legs of 3 and 6, the Pythagorean Theorem will help us figure out the third side (which is what we're asked to solve for): 3^2 + 6^2 = C^2 9 + 36 = 45 45 = C^2 (Root45) = C (Root45) = 3(Root5) Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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22 Apr 2017, 13:36
Y=X/2+1 Therefore Slope=1/2 Y intercept=1 i.e (0,1) X intercept=x i.e. (x,0)
These are two points on the line "m"
Hence the slope= 10/0x=1/2 Solving the above equation yields x=2
There fore x intercept=(2,0) Line m would intesect line x=4 at a point with co ordinate (4,y)
Consider two points on the same line"m" (4,y) and (0,1) y1/40=1/2 Solving it gives the value of y=3 (4,3) is the point at which Line m intersects line x=4
Horizontal Distance from (2,0) to (4,3) is 6 units Vertical distance between (4,0) and (4,3) is 3 units
Apply pytha theorem 6^2+3^2=H^2 root 45= h^2 h=3 root 5
This is the distance



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Knowing vertices (2,0) & (4,3), distance between two vertices is sqrt ((x2 x1)^2 + (y2y1)^2))
Thus sqrt(6^2+3^2) Sqrt(45) 3*sqrt(5)
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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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13 May 2017, 13:07
EMPOWERgmatRichC wrote: Hi reto, To answer this question, we'll first need to figure out the two coordinates involved.... Given the line Y = (X/2) + 1 The Xintercept will be the point at which Y = 0 0 = (X/2) + 1 1 = X/2 X = 2 So the first coordinate is (2, 0) The second coordinate is where X = 4.... Y = (4/2) + 1 Y = 3 So the second coordinate is (4, 3) The diagonal line segment that exists between these two points is the hypotenuse of a right triangle. That triangle has a base of 6 and a height of 3. So, with legs of 3 and 6, the Pythagorean Theorem will help us figure out the third side (which is what we're asked to solve for): 3^2 + 6^2 = C^2 9 + 36 = 45 45 = C^2 (Root45) = C (Root45) = 3(Root5) Final Answer: GMAT assassins aren't born, they're made, Rich Hi.. Why the answer is not 6? It is not mentioned in the question that we need to find the shortest distance.



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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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13 May 2017, 18:49
Hi nishantt7, The question asks us to determine the distance between two points on a graph, so there's only one answer  the actual distance between them (and that distance IS 3(root5)). "6" is the difference between the Xcoordinate of the two points, but that is NOT what the question asked for. GMAT assassins aren't born, they're made, Rich
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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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14 May 2017, 00:58
EMPOWERgmatRichC wrote: Hi nishantt7,
The question asks us to determine the distance between two points on a graph, so there's only one answer  the actual distance between them (and that distance IS 3(root5)). "6" is the difference between the Xcoordinate of the two points, but that is NOT what the question asked for.
GMAT assassins aren't born, they're made, Rich Thanks a lot. Got it. It asked for distance between (2,0) and (4,3). It is not horizontal or vertical distance.



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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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16 May 2017, 00:59
coordinates of point where m intersects with x axis is (2,0) by putting y=0 in equation of line m;
coordinates of point where m intersects with line x=4 is (4,3) by putting x=4 in eq of line m;
now find the distance between these 2 points (2,0) and (4,3) using distance formula = sqrt ( (4+2)^2 + (30)^2 ) = sqrt (45) = 3sqrt5
Hence OPTION D is correct



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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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Re: The equation of line m is y=x/2+1. What is the distance from the xint
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