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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
Even graphically, how do we find the respective end points from the mid point each corresponding to a length of \(\sqrt{52}/2\) as with the case of the example in the question?
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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
Sorry, I've edited this as my math was incorrect.

It's possible as MacFauz already pointed out.

Essentially, you want to use the Pythagorean Theorem.
You get \(x^2 + y^2 = (\sqrt{52}/2)^2.\)
You can then substitute \(y\) (just the slope part!) and get \(x^2 + (\frac{3}{2}x)^2 = 13\).
Solve for this and you get \(x = 2\). Plug this back into \(y = \frac{3}{2}x\) and you get \(y = 3\).
So now the two vertices are at \((3-2,4-3)\) and \((3+2,4+3)\).

Answer: \((1,1)\) and \((5,7)\)

Originally posted by farful on 13 Dec 2013, 06:33.
Last edited by farful on 13 Dec 2013, 06:52, edited 1 time in total.
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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
Thanks for the explanation farful... BTW, I do agree that this might not be a PS question but could very well be a DS question...
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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
Thanks for explaining. I'm little unclear on why would we substitute just the slope part in the equation and not the intercept?
In this problem, it does work since the y intercept falls beyond one of our end point. Is it always safe to ignore the y intercept?

BTW, I pulled pieces from a PS question to make it sound this way that I thought 'what if this were a DS question'.

farful wrote:
Sorry, I've edited this as my math was incorrect.

It's possible as MacFauz already pointed out.

Essentially, you want to use the Pythagorean Theorem.
You get \(x^2 + y^2 = (\sqrt{52}/2)^2.\)
You can then substitute \(y\) (just the slope part!) and get \(x^2 + (\frac{3}{2}x)^2 = 13\).
Solve for this and you get \(x = 2\). Plug this back into \(y = \frac{3}{2}x\) and you get \(y = 3\).
So now the two vertices are at \((3-2,4-3)\) and \((3+2,4+3)\).

Answer: \((1,1)\) and \((5,7)\)
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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
Sorry, I'm not really sure of the best way to explain this.

Let's take an easy example of y = x + 2. From the point (1,3) you want to know the point if you travel \(\sqrt{2}\) above the point (1,3). The answer is that you need to move 1 in the x direction, and 1 in the y direction. In fact, it doesn't matter where the intercept is (or the midpoint), if you want to travel \(\sqrt{2}\) along a y = x + k line, you will always travel 1 in the x direction and 1 in the y direction.
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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
**Edited the part highlighted in bold**
Great post Karishma! Very insightful. +1 K

So going by this method, it appears that we will be able to find the end points of a line segment just by knowing the mid point and [either the length of line segment or the equation of the line]?

VeritasPrepKarishma wrote:
mniyer wrote:
Hi,
Is it possible to find the vertices of a straight line with the following information?

1) we know the equation of the line
2) we know the length of the line
3) we know the mid-point of the line

For eg: Is it possible to find the end-points of the line?

1) \(y = \frac{3}{2}x - \frac{1}{2}\)
2) length = \(\sqrt{52}\)
3) mid point of the line is (3,4)


You don't have end points of a line. A line extends indefinitely in both directions. You have end points of a line segment.

And yes, you can easily find the end point co-ordinates. But before you do, you need to understand a few things.
The equation of a line is given as y = mx + c
m is the slope of the line. The slope means the change in y coordinate for a unit change in x coordinate. The equation of the line here is y = (3/2)x - 1/2. So for every 1 unit increase in x coordinate, y increases by 1.5 units. Similarly for every 1 unit decrease in x coordinate, y decreases by 1.5 units.

Now, according to the given information, this is what the line segment looks like:
Attachment:
Ques3.jpg


Since the length of the line segment is \(\sqrt{52}\), it must be \(\sqrt{52}/2 = \sqrt{13}\) on both sides of the mid point.

Now we need the length of the red arrow and blue arrow to get the coordinates. Since the slope of the line segment is 1.5, it means y coordinate changes by 3 units for every change of 2 units in x coordinate (got rid of the decimal). So length of blue line:length of red line = 3:2
Using pythagorean theorem, \((3a)^2 + (2a)^2 = \sqrt{13}^2\)
a = 1 or -1
So length of blue line is 3 units and length of red line is 2 units.
We obtain the coordinates as (3+2, 4+3) and (3 - 2, 4-3) i.e. (5, 7) and (1, 1)
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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
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mniyer wrote:
**Edited the part highlighted in bold**
Great post Karishma! Very insightful. +1 K

So going by this method, it appears that we will be able to find the end points of a line segment just by knowing the mid point and [either the length of line segment or the equation of the line]?



No, along with the mid point, you will need the length of the segment and the slope of the line (given by the equation or separately). Once you have the mid point and the length of the segment, you know you have to go up and down some distance from the mid point but the question is in which direction? You need to know the slop of the line too. I have used the slope of the line which is 3/2 here. It tells us that we move 3 units up for every 2 units right. So this specifies the direction of the line.
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Re: Finding the vertex of a line using eqn, length, midpoint [#permalink]
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