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In the x y-coordinate plane

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by sanju09 » Wed Sep 15, 2010 9:48 pm
In the x y-coordinate plane, the points (a, b) and (a + k, b - 3) are on the line defined by y = 2 x - 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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by limestone » Thu Sep 16, 2010 1:02 am
Point (a,b) is on the line y = 2x -5, then b = 2a -5
Point (a+k, b-3) is also on that line then: b-3 = 2(a+k) -5 or b -3 = 2a + 2k - 5
Then b = 2a + 2k -2
Subtract the second bold equation to the first one:
2a + 2k - 2 - (2a -5) = b- b = 0
2k + 3 = 0
then k =-3/2
Pick C.
Last edited by limestone on Thu Sep 16, 2010 5:16 am, edited 1 time in total.
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by Geva@EconomistGMAT » Thu Sep 16, 2010 1:10 am
Alternative solution using slope = (y2-y1 / x2-x1) and the fact the the slope of the line is 2 (the coefficient 'a' for x in the equation y=ax+b)

(b-3) - b / (a+k) - a = -3 / k, and that equals 2, so
-3/k = 2
k=-3/2.

Pick C.
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by Neo Gates » Thu Sep 16, 2010 2:05 am
sanju09 wrote:In the x y-coordinate plane, the points (a, b) and (a + k, b - 3) are on the line defined by y = 2 x - 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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slope of the line = [(b-3) - b] / [(a+k)-a] = -3/k

form the eq.( y = 2 x - 5) slope = 2

so, -3/k=2
k= -3/2
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by brijesh » Thu Sep 16, 2010 2:12 am
sanju09 wrote:In the x y-coordinate plane, the points (a, b) and (a + k, b - 3) are on the line defined by y = 2 x - 5. What is the value of k?
(A) -5/2
(B) -5/3
(C) -3/2
(D) -2/3
(E) -2/5


C- -3/2
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by sanju09 » Thu Sep 16, 2010 2:20 am
limestone wrote:Point (a,b) is on the line y = 2x -5, then b = 2a -5
Point (a+k, b-3) is also on that line then: b-3 = 2(a+k) -5 or b -3 = 2a + 2k - 5
Then b = 2a + 2k -8
From the 2 bold equations, 2k - 8 must be equal to -5
or 2k = -5 +8 = -3
k =-3/2
Pick C.
How do we get k = -3/2 from 2 a - 5 = 2 a + 2 k - 8?
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by limestone » Thu Sep 16, 2010 5:12 am
Well, I made a mistake here. Sorry guys. b is not equal to 2a + 2k - 8 as I stated above

b-3 = 2(a+k) -5 or b = 2a + 2k -5 +3 or b = 2a + 2k -2
Also : b = 2a -5
then 2a -5 = 2a + 2k -2 or 2k = -3 or k=-3/2

I have also edited my answer above.
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by sbeagol1000 » Wed Mar 13, 2013 10:04 am
This is what I did

(a,b)=(x,y)

x=a+k
y=b-3

original equation is y=2x-5,so...

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

Then I started solving the equation for k

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

b+2=2(a+k)

(b+2)/2 = a + k

((b+2)/2)-a = k

combine everything into one fraction

(b+2-2a)/2=k

I picked a coordinate (0,-5) that would satisfy y=2x-5 and plugged it in

(-5+2-2*0)/2 = k

(-5+2-0)/2 = k

-3/2 = k

*looks like a long process but when you are writing it down it is much faster than it seems*
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