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The equation of line m is y=x/2+1. What is the distance from the x-int

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The equation of line m is y=x/2+1. What is the distance from the x-int  [#permalink]

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New post 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 x-intercept 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 x-int  [#permalink]

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New post 12 May 2015, 22:18
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Hi reto,

To answer this question, we'll first need to figure out the two co-ordinates involved....

Given the line Y = (X/2) + 1

The X-intercept will be the point at which Y = 0

0 = (X/2) + 1
-1 = X/2
X = -2

So the first co-ordinate is (-2, 0)

The second co-ordinate is where X = 4....

Y = (4/2) + 1
Y = 3

So the second co-ordinate 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:

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Re: The equation of line m is y=x/2+1. What is the distance from the x-int  [#permalink]

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New post 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= 1-0/0-x=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)
y-1/4-0=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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New post 22 Apr 2017, 17:53
Knowing vertices (-2,0) & (4,3), distance between two vertices is sqrt ((x2- x1)^2 + (y2-y1)^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 x-int  [#permalink]

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New post 13 May 2017, 13:07
EMPOWERgmatRichC wrote:
Hi reto,

To answer this question, we'll first need to figure out the two co-ordinates involved....

Given the line Y = (X/2) + 1

The X-intercept will be the point at which Y = 0

0 = (X/2) + 1
-1 = X/2
X = -2

So the first co-ordinate is (-2, 0)

The second co-ordinate is where X = 4....

Y = (4/2) + 1
Y = 3

So the second co-ordinate 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 x-int  [#permalink]

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New post 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 X-coordinate of the two points, but that is NOT what the question asked for.

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Re: The equation of line m is y=x/2+1. What is the distance from the x-int  [#permalink]

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New post 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 X-coordinate 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 x-int  [#permalink]

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New post 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 + (3-0)^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 x-int  [#permalink]

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Re: The equation of line m is y=x/2+1. What is the distance from the x-int   [#permalink] 21 Mar 2019, 17:03
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