Key Concept: If a point lies ON a line, then the coordinates (x and y) of that point must SATISFY the equation of that line.

Given equation: x = 3y - 7
One point ON the line is (a, b)
So, we can write: a = 3b - 7

Another point ON the line is (a + 3, b + k)
So, we can write: a + 3 = 3(b + k) - 7
Expand: a + 3 = 3b + 3k - 7
Subtract 3 from both sides to get: a = 3b + 3k - 10

We now two equations:
a = 3b + 3k - 10
a = 3b - 7

Subtract the bottom equation from the top equation to get: 0 = 3k - 3
Add 3 to both sides: 3 = 3k
Solve: k = 1

Answer: D

Cheers,
Brent
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given : y = 1/3x + 7/3

Slope => k/3 = 1/3 i.e K =1

x=3*y-7
convert to (this step is not necessary but I do it out of habit)
y=(x-7)/3

substitute a and b for x and y
1. b=(a-7)/3

b+k=((a+3)-7)/3
substitute equation 1 for b
(a-7)/3+k=((a+3)-7)/3
(a-7)/3+k=(a-4)/3
a-7+3k=a-4
3k=3
k=1

D

x = 3y - 7
y = 1/3 (x) + 7/3
[(b+k) - b]/ [(a+3) - a] = 1/3
k/3 = 1/3
k = 1

D.

slope = (y1-y2)/(x1-x2) = (b+k-b) / (a+3-a) = k/3
y = x/3 + 7/3, so slope = 1/3

k/3 = 1/3
k = 1

Substitute a & b in place of x & y resp... in the eqn.
a-3b+7=0------(i)

Substitute a+3 & b+k in the place of x & y resp...we'll get
a+3=3(b+k)-7 or,
a-3b-3k+10=0-------(ii)

points on the same line will satisfy the equation so ,

equating (i)&(ii)
a-3b-3k+10=a-3b+7
k=1

Ans- D

Or note here itself that a - 3b + 7 = 0 so 3 - 3k = 0 giving you k = 1
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My answer is (D).

Here's my approach:

We are given two points: (a,b) and (a+3,b+k)

We are given an equation of the line: x = 3y - 7

Next step is to convert the equation of the line into slope-intercept form. We will have Y = x/3 + 7/3. Don't forget this step because you might fall into the trap and decide that the slope of the line is 3. This makes the question very GMAT-esque.

So the slope of the line is 1/3

Now we know that the equation of the slope of the line is given by:

slope = (y2 - y1)/(x2 - x1) ---> Remember, the slope is just "rise" over "run."

That's why we have:

1/3 = [ (b+k) - b ] / [ (a+3) - a]

The two b's will cancel each other in the numerator and so will the two a's in the denominator

We will get 1/3 = k / 3

so 3 / 3 = k

k = 1

Hi,

Slope of the line x=3y-7 is 1/3
so, we can equate the slope of the line to the slope of the points = \(\frac {y_2-y_1}{x_2-x_1}\)
or \(\frac {(b+k)-(b)}{(a+3)-a} = \frac 13\)
or \(\frac k3 = \frac 13\)
or k=1,

Answer (D),

Regards,

x=3*y-7 is not y=(x-7)/3
instead it is y=(x+7)/3 or y=(1/3)x+(7/3)
since we know (a,b) and (a+3,b+k) belong to the same line, they must have the same slope, 1/3.
[(b+k)-b]/[(a+3)-a]=1/3
k/3=1/3
k=1

Hi All,

This question can be solved by TESTing VALUES:

We're given the equation of a line (X = 3Y - 7) and we're told that two points (A, B) and (A+3, B+K) are on this line. We're asked for the value of K.

In graphing questions, it sometimes helps to "visualize" the line better if you write the equation in "slope-intercept" format:

X = 3Y - 7

3Y = X + 7
Y = X/3 + 7/3

For the first co-ordinate, let's try to keep things simple...
X = 0
Y = 7/3

So...
A = 0
B = 7/3

For the second co-ordinate, we have ADD 3 to X....
X = 3
Y = 10/3

So....
A+3 = 3
B+K = 10/3

We know from the first co-ordinate that B = 7/3, so K = 3/3 = 1

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Our equation is x=3y-7
Lets quickly make it a point intercept form ==> 3y-7=x==>3y=x+7==>\(y=\frac{x}{3}+\frac{7}{3}\)
Now we can see the coefficient of x is \(\frac{1}{3}\), which by definition is the slope
and we know slope =\(\frac{y2-y1}{x2-x1}\)==> \(\frac{1}{3}=\frac{b+k-b}{a+3-a}\)==> 1/3=k/3
therefore k=1

Answer is B
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Recall that an ordered pair represents a pair of x and y coordinates. Substituting the values from the first ordered pair (a,b) into the equation, we can create the following equation:

a = 3b - 7

Substituting the values from the second ordered pair for x and y into the same equation, we have:

a + 3 = 3(b + k) - 7 → a + 3 = 3b + 3k - 7

If we subtract the first equation from the second, we have:

3 = 3k

1 = k

Answer: D
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Hi Rich, I know the "7/3" came from rewriting to slope-intercept form, but how did you get from "B = 7/3" to "B+K = 10/3"? I know we added 3 to X, but are we adding 3 to Y too?

Hi OCDianaOC,

Both co-ordinates have to 'fit' the equation Y = X/3 + 7/3

The first co-ordinate is (A, B).... and I TESTed VALUES and used (0, 7/3) to define that co-ordinate. Remember: A = 0 and B = 7/3

The second co-ordinate is (A+3, B+K).... notice how that's the SAME A and B from the first co-ordinate. Thus, we have to add 3 to A (so 3+0 = 3) and plug in X=3 into the equation to get the value of the Y....

When X=3....
Y = X/3 + 7/3
Y = (3/3) + 7/3)
Y = 10/3

Since the second co-ordinate is (A+3, B+K), our prior work makes the co-ordinate (3, 10/3). From the prior work, we see that B = 7/3...
B + K = 10/3
(7/3) + K = 10/3
K = 10/3 - 7/3 = 3/3 = 1

GMAT assassins aren't born, they're made,
Rich
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How did you figure out that slope is 1/3 from here Y = x/3 + 7/3 slope is x/3 but how should i know value of X ?

\(x = 3y - 7\)

Rewrite in slope-intercept form
\(y = mx + b\)
m = slope, b = y-intercept, thus:
\(3y = x + 7\) =>
\(y = \frac{1}{3}x + \frac{7}{3}\)

Slope = \(\frac{1}{3}\)

Slope is also \(\frac{rise}{run}=\frac{(y_2 - y_1)}{(x_2 - x_1)}\)

We have x-and y-coordinates for two points:
(a,b) and (a+3,b+k)

Set the slope equation equal to the slope value

\(\frac{(b+k)-b}{(a+3)-a}=\frac{1}{3}\)

\(\frac{b+k-b}{a+3-a}=\frac{1}{3}\)

\(\frac{k}{3}=\frac{1}{3}\)

\(3k = 3\)

\(k = 1\)

Answer D
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The equation of a line is
y = mx + c
where m is the slope and c is the y-intercept.
So the equation looks like this
y = 2x + 4 (m = 2, c = 4)
y = x/3 + 5 (m = 1/3, c = 5)
etc

For more, check out this post:
http://www.veritasprep.com/blog/2010/12 ... he-graphs/
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another solution since x=3y-7 is increasing line ie if x increases then y increasing(x will increase by y's constant y will increase by x's constant)
hence D