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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # ln the coordinate plane, line k passes through the origin

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Intern  B
Joined: 07 Apr 2019
Posts: 31
Re: ln the coordinate plane, line k passes through the origin  [#permalink]

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generis wrote:
CyberStein wrote:
Quote:
y = 2x is the equation of the line - fine.

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.

Think of them instead as points (3, a) and (b, 4). You need to find a + b.
You will get 10 = 2a + b by equating it to slope but how will you solve for a + b?

The equation of the line is y = 2x
So a = 2*3 = 6
Also, 4 = 2*b
b = 2

So a + b = 6 + 2 = 8

I got the question at hand right (C), but I am uncomfortable as to how I got it right -- it felt like guessing.

here is what I did, step by step:

1. y = mx + b
2. y = 2x + 0 given slope = 2, and the origin
3. y = 2x
After step three, I dead ended and did not know how to use "y = 2x" to solve for the missing points.

So, I just looked at the problem from a rise over run POV

2 = rise / run

2 = y-4 / 3-x <--- right here, I essentially guessed by plugging in numbers until the equation worked.
2 = (6) - 4 / 3 - (2) <--- I plugged in 6 for Y and 2 for X because I knew it would give me 2; however, I know that this way is essientially guessing.

What is a simplified solution to the problem, following the steps I followed?

CyberStein
You had it! I think you just forgot that you already found the relationship between y and x (in the equation) -- and, once you have the linear function in the form of the equation, you can find one coordinate if you have the other.

The missing x and y have nothing to do with one another except that they are on the same line. The question is asking for one more trivial step after you do the work: can you track correctly and add?

All you need to do: use the line equation you found, take one pair of coordinates at a time, and plug in the given value to find the missing one.

For (3,y). We know THIS x is 3. What is its matching y-coordinate? Plug in the given x-value to find the y-value for this pair. We need the "missing" y in (3,y).

y = 2x
Given: x = 3. Plug it in.
y = (2)(3)
y = 6
y equals 6 when x equals 3. The pair is now (3, 6).
If you were graphing, and went to x equals 3, you would go up to y equals 6.

Now the other pair. We are given (x,4). We need x.

y = 2x
Given: y = 4. Plug it in.
4 = 2x
4/2 = x
x = 2
For this pair we have (2,4).

This pair is a little odd because we are used to thinking of y as a function of x, and not the other way around. But we already have the relationship between y and x. It's in the equation you wrote.

The question at the end is trivial, meaning, it has nothing to do with the relationship between the x and the y -- except what a randomly picked x-value and a randomly picked y value, from that line, add up to.

The question is, my rewrite: "[Now that you have figured out the line equation, and figured out particular values for missing coordinates from two different pairs], what does [that] x + [that] y equal?"

We got x equals 2 on the one hand, and y equals 6, on the other hand. Add them.

2 + 6 = 8

Maybe you were thinking too hard, that's all. I should all have such problems! I couldn't quite tell where you were going wrong, so I guessed. Does that help?

Hi generis (other experts as well)

why is this happening ?

with (3,y) and (x,4) we know the slope is +2. If I use slope formula y2-y1/xx2-x1 ---- y-4/3-x = 2 [ which I think we can use given two points, i'm not concerned about getting the value of x+y at this point]
this gives us an equation ==> y = -2x + 10. I'm wondering how this is possible, because in the slope intercept form of this equation the slope is turning out to be -2.

I can see y = 2x is the right way ... but for once if i consider y2-y1/x2-x1 = m, the slope intercept form of the equation should give me the slope with the same sign correct? while here its coming out to be -2. Can you help me understand why this is happening?

Further, for those who went ahead with this method (mentioned above in the thread) I think this is the point from which things went haywire for them (when the equation is not consistent with the desired result, the calculation based on that would definitely turn out improper). Seeking feedback. Re: ln the coordinate plane, line k passes through the origin   [#permalink] 08 Dec 2019, 05:36

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