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

Question

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

Bunuel · Answer

Similar questions: https://gmatclub.com/forum/in-the-xy-co ... 35012.html https://gmatclub.com/forum/in-the-xy-co ... 55642.html

DavidTutorexamPAL · Answer

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.

ccooley · Answer

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

rahul16singh28 · Answer

b = 2a - 5, 
 b - 3 = 2(a+k) - 5... K = -3/2

AkshdeepS · Answer

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}

bumpbot · Answer

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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