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In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are

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In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are [#permalink]

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New post 21 Jan 2018, 03:31
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In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are on the line defined by y = 2x – 5. What is the value of k ?

(A) -5/2
(B) -5/3
(C) -3/2
(D) -2/3
(E) -2/5

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Re: In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are [#permalink]

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New post 21 Jan 2018, 03:32
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Bunuel wrote:
In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are on the line defined by y = 2x – 5. What is the value of k ?

(A) -5/2
(B) -5/3
(C) -3/2
(D) -2/3
(E) -2/5


Similar questions:
https://gmatclub.com/forum/in-the-xy-co ... 35012.html
https://gmatclub.com/forum/in-the-xy-co ... 55642.html
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are [#permalink]

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New post 21 Jan 2018, 12:04
Bunuel wrote:
In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are on the line defined by y = 2x – 5. What is the value of k ?

(A) -5/2
(B) -5/3
(C) -3/2
(D) -2/3
(E) -2/5


Since there is a specific rule which leads to a straightforward calculation, we'll just use it.
This is a Precise approach.

When we're given two points, we can calculate the slope of the line through them with (y2 - y1)/(x2 - x1).
So, 2 = (b - 3 - b)/(a + k - a) = -3/k.
Then k = -3/2.

(C) is our answer.
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Re: In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are [#permalink]

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New post 21 Jan 2018, 12:55
Bunuel wrote:
In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are on the line defined by y = 2x – 5. What is the value of k ?

(A) -5/2
(B) -5/3
(C) -3/2
(D) -2/3
(E) -2/5


Points with variables, plus a line equation, makes me think that this is just an algebra problem dressed up in geometry clothing. :)

After all, if a point is on a line, and you have the equation for that line, all that means is that you can plug in the coordinates of the point, to the equation for the line, and you get a true statement.

So, we can plug in both points and get correct equations:

b = 2a - 5
b-3 = 2(a+k) - 5

Simplify the second equation:

b - 3 = 2a + 2k - 5

b = 2a + 2k - 2

Since both equations now tell you the value of b, you can set them equal to each other:

2a - 5 = 2a + 2k - 2
-5 = 2k - 2
-3 = 2k
k = -3/2
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Re: In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are [#permalink]

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New post 22 Jan 2018, 23:22
Bunuel wrote:
In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are on the line defined by y = 2x – 5. What is the value of k ?

(A) -5/2
(B) -5/3
(C) -3/2
(D) -2/3
(E) -2/5


\(b = 2a - 5,\)
\(b - 3 = 2(a+k) - 5... K = -3/2\)
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Re: In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are [#permalink]

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New post 23 Apr 2018, 05:17
Bunuel wrote:
In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are on the line defined by y = 2x – 5. What is the value of k ?

(A) -5/2
(B) -5/3
(C) -3/2
(D) -2/3
(E) -2/5


We have two points and an equation of a straight line.

Points : (a,b) & (a+k), (b-3)

Equation of a straight line :\(y = 2x – 5\), in which 2 is slope and -5 is y - intercept

Formula says, \(m = \frac{Y2 - Y1}{X2 - X1}\)

\(2 = \frac{b - 3 - b}{a + k - a}\)

\(2 = \frac{- 3}{k}\)

\(k = \frac{-3}{2}\)
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Re: In the xy-coordinate plane, the points (a, b ) and (a + k, b – 3) are   [#permalink] 23 Apr 2018, 05:17
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