Line x = 2y + 5

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kartikshah: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sun Jul 15, 2012 12:03 pmFollowed by: 1 member; Location: New York, USA; Thanked: 34 times

by kartikshah » Mon Jul 16, 2012 3:36 am

In a rectangular coordinate system, if the line x = 2y + 5 passes through points (m,n) and (m+2, n+p), what is the value of p?
1. -2
2. 0
3. Â½
4. 1
5. 2

Please help!
Source: Princeton Review Practice Test 10

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Source: Beat The GMAT — Problem Solving | Mon Jul 16, 2012 3:36 am

eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Mon Jul 16, 2012 5:26 am

kartikshah wrote:In a rectangular coordinate system, if the line x = 2y + 5 passes through points (m,n) and (m+2, n+p), what is the value of p?
1. -2
2. 0
3. Â½
4. 1
5. 2

Please help!
Source: Princeton Review Practice Test 10

Easiest way is to compare slopes:

x = 2y + 5 => y = x/2 -5/2
Recall that slope of y = mx+c is m. Here m = 1/2

Now slope of line segment joining (m,n) with (m+2,n+p) = (n+p-n)/(m+2-m) = p/2
Since this is part of the same line, slopes are equal. Hence p/2 = 1/2 => p = 1.

option 4 is correct.

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kartikshah: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sun Jul 15, 2012 12:03 pmFollowed by: 1 member; Location: New York, USA; Thanked: 34 times

by kartikshah » Mon Jul 16, 2012 8:11 am

Eagleeye: I understood how you derived the slope of the line. But did not follow your calculations thereafter. Please can you explain stepwise?
I'm not that great with Coordinate Geometry.

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mekoner: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sun Jul 15, 2012 9:43 pm; Thanked: 1 times

by mekoner » Mon Jul 16, 2012 12:14 pm

kartikshah wrote:Eagleeye: I understood how you derived the slope of the line. But did not follow your calculations thereafter. Please can you explain stepwise?
I'm not that great with Coordinate Geometry.

The calculations after the slope are just simply Rise/Run.

We have:
(X1, Y1) = (m,n)
(X2,Y2) = (m+2, n+p)

We then simply solve by Y2-Y1 / X2-X1.

----------
[quote="eagleeye"Easiest way is to compare slopes:

x = 2y + 5 => y = x/2 -5/2
Recall that slope of y = mx+c is m. Here m = 1/2

Now slope of line segment joining (m,n) with (m+2,n+p) = (n+p-n)/(m+2-m) = p/2
Since this is part of the same line, slopes are equal. Hence p/2 = 1/2 => p = 1. [/quote]

My question is:
"Since this is part of the same line", meaning that x = 2y + 5 = y = x/2 -5/2, right? And, should we always re-write the formula in the Y=mx+b form?

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Mon Jul 16, 2012 5:59 pm

mekoner wrote: My question is:
"Since this is part of the same line", meaning that x = 2y + 5 = y = x/2 -5/2, right? And, should we always re-write the formula in the Y=mx+b form?

a. Yes, that means the line segment is a part of x = 2y+5.
b. Equation of a line can be written in different forms. Once you have enough experience, you can find out the slope without actually using the form. However, writing things out in a recognized form prevents mistakes. At the same time, if you can calculate the slopes mentally or just by observation, you can save computational time. So, to finally answer your question, I would suggest getting familiar with the various forms of equation of a line before doing mental calculations.

Here are a few forms of the top of my head:

1. x/a + y/b = 1 (where a is x intercept and b is y intercept).
2. y = mx + b (where m is the slope and b is the y intercept)
3. (y-y1) = m(x-x1) (where m is the slope and the line passes through (x1,y1)
4. (y-y1) = (x-x1)*(y2-y1)/(x2-x1) (where the line passes through (x1,y1) and (x2,y2)

From comparing the various forms we can see that:

slope = m = -b/a = (y2-y1)/(x2-x1)
and x-intercept = a = -b/m = (x1*y2-x2*y1)/(y2-y1)

Cheers!

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das.ashmita: Master | Next Rank: 500 Posts; Posts: 118; Joined: Mon May 21, 2012 10:07 pmFollowed by: 4 members; Thanked: 23 times

by das.ashmita » Mon Jul 16, 2012 7:59 pm

Here is Another approach

By substitution
x=2y+5

For pt(m,n):
m=2n+5 => 2n = m-5...(1)

For pt(m+2,n+2):
and m+2 = 2(n+p) +5
=>m+2 = m-5+2p+5 (by substituting eqn 1)
=>q=1

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