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# In the coordinate plane, points (x, 1) and (10, y) are on line k. If

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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
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We know the equation of any line is y=mx+c

Where m = slope and c = y intercept (point where line crosses y axis)

Since line passes thru origin then y intercept (i.e. C) must be zero

we also know m = slope = 1/2

so the equation becomes y=$$\frac{x}{2}$$

we are given two points that are on line (x,1) (10,y)

Plug second point (10,y) in to the equation y=$$\frac{10}{2}$$ --------> y=5

Plug First point (x,1) in to the equation 1=$$\frac{x}{2}$$ -----------> x=2

x+y = 7

Choice B

Regards,

Narenn
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
manishuol wrote:
In the coordinate plane, points (x, 1) and (10, y) are on line k. If line k passes through the origin and has slope 1/2, then x + y =

(A) 4.5
(B) 7
(C) 8
(D) 11
(E) 12

Equation of a line passing through origin and having a slope m is given by $$y = mx$$

Hence the equation of the given line is $$y = 0.5 * x$$

This gives the values of x and y as 2 and 5 respectively.
Hence x + y = 7

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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
My approach was to apply the rise/run=slope rule

so,

y-1/10-x = 1/2
2y-2=10-x
2y+x=12, substitute with one of the given points (x,1)
2(1) +x=12, x=10, y=1, x+y=11

can someone please tell me why this approach doesn't work?
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
SaraLotfy wrote:
My approach was to apply the rise/run=slope rule

so,

y-1/10-x = 1/2
2y-2=10-x
2y+x=12, substitute with one of the given points (x,1)
2(1) +x=12, x=10, y=1, x+y=11

can someone please tell me why this approach doesn't work?

IMO what you have obtained (2y+x=12) is the algebraic equation, but it is not the equation of a line.

Equation of line is described as y=mx+c ------> where 'm' is the slope of a line and 'c' is the 'y' intercept.

Per the equation 2y+x=12

$$m=slope=-\frac{1}{2}$$ (which is not correct. The slope is $$\frac{1}{2}$$)

c=y intercept = 6 (This is also not correct. We know that line is passing thru origin, so its y intercept (the y value of point where line crosses 'y' axis) must be zero.
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
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Official Explanation

Answer: B We're given the slope of a line and one point on the line (the origin: 0,0). From this, we can determine every other point on the line. The most direct way to find x and y involves solving for each individually.

We know that slope (1/2, in this case), is equal to the change in y divided by the change in x. Since we know that the line passes through (0,0) and (x,1), we can solve as follows:

1/2 = (1 – 0)/(x – 0)
1/2 = 1/x
x = 2

Use the same approach to solve for y:

1/2 = (y – 0)/(10 – 0)
1/2 = y/10
y = 5

Thus, x + y = 2 + 5 = 7, choice (B).
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
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mau5 wrote:
avohden wrote:
In the coordinate plane, points (x, 1) and (10, y) are on line k. If line k passes
through the origin and has slope 1/2, then x + y =

(A) 4.5
(B) 7
(C) 8
(D) 11
(E) 12

As both the points are on the same line, and the line passes through the origin with a positive slope,
$$\frac{1-0}{x-0} = \frac{1}{2}$$ and $$\frac{y-0}{10-0} = \frac{1}{2}$$

Hence, x=2 and y=5

Thus, x+y = 7

B.

IMO more like 550.

One can only check the coordiantes and solve

See, we are given that y = 1/2x

So first coordinate since y =1 then x = 2
Second coordinate since x = 1= then y=5

So add em up 2+5 = 7

B

Hope its clear

Cheers
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
Hi All,

This question has a great "brute force" element to it. Since we know that the line passes through the Origin (0,0) and has a slope of 1/2, we can "map out" as many co-ordinates as we need to to answer the given question.

Since the slope is 1/2, for every increase of 2 in the X-coordinate we have an increase of 1 in the Y-coordinate:

(0,0)
(2,1)
(4,2)
(6,3)
(8,4)
(10, 5)

We're told that (X,1) and (10,Y) are on the line. We're asked for the value of X+Y....

From our list of co-ordinates, we can see that X = 2 and Y = 5...

X+Y = 2+5 = 7

GMAT assassins aren't born, they're made,
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
Slope m = (y2 - y1) / (x2 - x1) ---formula equation

there are three points on the line (0,0) , (x,1) and (10,y).

given m = 1/2.

first take (0,0) , (x,1) and m=1/2 and put these in formula equation and we get
1/2 = 1-0 / x-0 hence x=2

first take (0,0) , (10,y) and m=1/2 and put these in formula equation and we get
1/2 = y-0 / 10-0 hence y=5

x+y = 5+2 = 7.
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
manishuol wrote:
In the coordinate plane, points (x, 1) and (10, y) are on line k. If line k passes through the origin and has slope 1/2, then x + y =

(A) 4.5
(B) 7
(C) 8
(D) 11
(E) 12

Hi niks18 please let me know if my solution/ approach is correct ?

$$\frac{1-y}{x-10}=\frac{1}{2}$$ cross multiply

$$2-2y = x-10$$

$$x-10-2+2y$$

$$x+2y-12 = 0$$

$$2y= 12-x$$

$$y = \frac{-x}{2}+ 6$$

now since I know that slop is $$1/2$$ hence x = 1 and y intercept is 6

so x+y = 1+6 = 7

many thanks!
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
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dave13 wrote:
manishuol wrote:
In the coordinate plane, points (x, 1) and (10, y) are on line k. If line k passes through the origin and has slope 1/2, then x + y =

(A) 4.5
(B) 7
(C) 8
(D) 11
(E) 12

Hi niks18 please let me know if my solution/ approach is correct ?

$$\frac{1-y}{x-10}=\frac{1}{2}$$ cross multiply

$$2-2y = x-10$$

$$x-10-2+2y$$

$$x+2y-12 = 0$$

$$2y= 12-x$$

$$y = \frac{-x}{2}+ 6$$

now since I know that slop is $$1/2$$ hence x = 1 and y intercept is 6

so x+y = 1+6 = 7

many thanks! :)

Hi dave13,

what you have done is used the formula to find slope and converted it to an algebraic equation which does not represent the equation of line.

equation of line is $$y=mx+c$$, where $$m$$ is slope of the line

as per your equation $$y=\frac{-1}{2}x+6$$, so here slope, $$m=\frac{-1}{2}$$ which is incorrect.

We know that the line passes through the origin so our equation should be

$$y=\frac{1}{2}x+c$$ and at origin we have (0,0)

so $$0=\frac{1}{2}*0+c => c=0$$ i.e y-intercept is 0 (as per your equation y intercept is 6 which is incorrect). If a line passes through origin it will not cut y-axis and hence there will be no intercept.

Hence equation of line will be $$y=\frac{1}{2}x$$

now at (x,1) we will have $$1=\frac{1}{2}x=>x=2$$

and at (10,y) we will have $$y=\frac{1}{2}*10 =>y=5$$

Hence $$x+y=2+5=7$$

There is an alternate method as well using only the formula to find slope. This method is also explained in earlier posts.
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
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manishuol wrote:
In the coordinate plane, points (x, 1) and (10, y) are on line k. If line k passes through the origin and has slope 1/2, then x + y =

(A) 4.5
(B) 7
(C) 8
(D) 11
(E) 12

Since the slope m of the line is ½ and the line passes through (x, 1) and the origin (0,0), we use the slope formula m = (y1 - y2)/(x1 - x2) and we have:

(1 - 0)/(x - 0) = 1/2

1/x = 1/2

x = 2

Similarly, using the points (10, y) and the origin, we can create the equation:

(y - 0)/(10 - 0) = 1/2

y/10 = 1/2

y = 5

Thus, x + y = 2 + 5 = 7.

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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
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