Should (2) read ´for some constant k' ?

If K is the constant, I'm thinking B

What we're trying to prove is ac = a^2 or a = c.

Statement 1: d = b +2 - doesn't tell us anything about the a vs. c relationship. INSUFFICIENT

Statement 2: I think we can rephrase L2's equation as y + k = ax + b based on this knowledge. Since k is constant, we know the lines increase at the same rate, it's just that one line has a higher y intercept (the difference being k).



I can't explain it much better, take some examples to visualize it.

Slope 1 = 1; Slope 2 = 2.

(1,1) (1,2)
(2,2) (2,4)
(3,3) (3,6)

This shows us that the k is increasing not constant. At x = 1, k = 1, at x = 2, k = 2 and so on.

Slope 1 = 1; slope 2 = 1/2

(1,1) (1,1/2) <- not a point
(2,2) (2,1)
(3,3) (3,3/2) < - not a point

So we have x values that don't work, and k also isn't constant. (it increases by 1/2 every time.

Slope 1 = 1; slope 2 = -1

(1,1) (1,-1)
(2,2) (2,-2)
(3,3) (3,-3)

We can see that this also has a non-constant k. The difference at point 1 is 2. The difference at point 2 is 4. and so on.

The only way we can keep K constant is to have the slopes equal.

kevincanspain wrote:Should (2) read ´for some constant k' ?

No. 2) does say "for some constant x."
Thanks.

lrambouskova wrote:If l1 and l2 are distinct lines in the xy coordinate system such that the equation for l1 is y=ax+b and the equation for l2 is y=cx+d, is ac=a² ?
1) d=b+2
2) For each point (x, y) on l1, there is a corresponding point (x, y+k) on l2 for some constant x.

Can someone explain this question for me? I am having a hard time understanding it. I do understand the y=mx+b formula, but I don't know how to solve this. Thank you.

Lenka.

line l1; y=ax+b;
l2; y=cx+d;

now to check whether ac=a^2; we need some sort of relationship between line l1 and l2;

1) d=b+2;

l1; y=ax+b; and l2; y=cx+d;
thus l2 become y=cx+b+2; as we cannot conclude anything about a and c, hence 1 alone is not sufficient to answer the question.

2)l1;y=ax+b; line l2=cx+d;
since for point (x,y) on line l1 there is corresponding point (x,y+k) on line l2; therefore we have;

y=ax+b;
(y-b)/a=x; also,
y+k=cx+d;
(y+k-d)/c =x;

now as x is constant, therefore
(y-b/a)=(y+k-d)/c; now for RHS to become equal to LHS we must have same numerator as well as same denominator; i.e. (y-b)=(y+k-d) and a=c;
y-b=y+k-d;
k=d-b;
and also as a=c; therefore two lines l1 and l2 must have equal slopes; hence 2 alone is sufficient to answer the question hence B

This could also be done through a simpler approach.

Given:
L1 : y = ax + b
L2 : y = cx + d

To find : ac=a^2 -> in other words are lines L1 & L2 parallel?
ac = aa --> a = c ?

Evaluating Statement 1

d = b + 2 -> This talks about y intercept and has no relevance on comparative slopes

Therefore Insufficient


Evaluating Statement 2

If you plot a line using test values you will find the lines are parallel to each other (K could be assumed to be anything 1,2,-1, etc)

(1,1) and (1,1+k)
(2,2) and (2,2+k)

Therefore sufficient