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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

Correct Answer is B
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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]
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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
J :)
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Re: In the coordinate plane, points (x, 1) and (10, y) are on line k. If [#permalink]
Expert Reply
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

Final Answer:

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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

Answer: 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

Answer: 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.

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