In the xy-coordinate system, if (m, n) and (m + 2, n + k) are two poin

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

In the xy-coordinate system, if (m, n) and (m + 2, n + k) are two points on the line with the equation x = 2y + 5, then k =

A. 1/2
B. 1
C. 2
D. 5/2
E. 4



Since (m, n) and (m + 2, n + k) are two points on the line with the equation x = 2y + 5 they should satisfy m=2n +5 and m+2 =2 *(n+k)+5.
By 1st equation we have m-2n=5 and by 2nd equation m-2n = 2k+3 ---> 5=2k+3 --->k=1.


The answer is, therefore, (B).

Hi All,

This question can be solved by TESTing VALUES.

We're given the equation X = 2Y + 5. While it's perfectly fine to graph this question in this format, many Test Takers find it easier to 'convert' any formulas into 'slope-intercept' format, so I'll do that here...

X = 2Y + 5
2Y = X - 5
Y = X/2 - 5/2

We're told that two points exist on this line: (m, n) and (m + 2, n + k). We're asked for the value of k.

For the first co-ordinate, let's TEST m=6...

Y = (6)/2 - 5/2 = 1/2
So n = 1/2

The first co-ordinate is (6, 1/2)

For the second co-ordinate, we already have the m and n....

(m+2, n+k) --> (6+2, 1/2+k) --> (8, 1/2 + k)

Y = (8)/2 - 5/2 = 3/2
so n+k = 3/2

We already know that n=1/2...

1/2 + k = 3/2
k = 2/2 = 1

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Theory: If two points are lying on a line, then they satisfy the equation of the line.

Putting (m,n) in the equation of the line,
m = 2n + 5 (i)

Putting (m + 2, n + k) in the equation of the line,
m + 2 = 2n + 2k + 5

On rearranging,
m = 2n + 2k +3 (ii)

Since both are the equations of the same line,
From (i) and (ii), we know that, 2k+ 3 = 5
Or k = 1.
Option B

we have been given X1, Y1 & X2 & Y2
and equation 2y=x-5
Y=x/2-5
hence m=1/2
now solve given co ordinate for m i.e=Y2-Y1/X2-X1
m=n+k-n/m+2-m will lead us k/2
Now solve for m=1/2=k/2 & will get m =1

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