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Homework Help: Need help with some algebra

  1. Mar 20, 2010 #1
    These are a few questions I couldn't do in my book, which are pretty important to know how to do :redface:

    1. Point, (h,k) lies on line y = x + 1 and is 5 units from point (0, 2), find all values of h and k after making two equations for them.

    My attempt was to make (sqrt (0-h)^2 + (2-k)^2)= 5 but didn't get anywhere

    2. solved

    3. P & Q are points of intersection at y/2 + x/3 = 1 which becomes (0, 2), (3, 0). Gradient of QR is 1/2, where R is point x-coordinate 2a, a>0

    The question is find y-coord of R in terms of a but more importantly how do I draw this up to work from?

    4. A triangle ABC with A (1,1) B (-1,4) gradient AB, AC and BC are -3m,3m and m, find value for m and coordinates of C and show that AC=2AB
    Last edited: Mar 20, 2010
  2. jcsd
  3. Mar 20, 2010 #2


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    Actually, you are getting somewhere.
    Notice how it says making two equations for them. Well, you've made one. You also know that the point (h,k) needs to be on the line y=x+1. What if you let y=k and x=h, then you can solve for both variables simultaneously :smile:

    Good question. I can't figure out which point P and Q are meant to be. Is P the y or x intercept? Lets assume R is the point (3,0) for now. Draw up your line y/2+x/3=1, then draw a line with gradient 1/2 going through Q.
    Choose any point on that line (easier if you choose it in the 1st quadrant) and label that R. The x-value is 2a.
    Now what is the equation of that line? which Q and R lie upon?

    Are you at least able to draw this on a diagram? It would be much easier to solve if you do that. First, notice that the gradient of AB is -3m, and that gradient is negative, so you're obviously dealing with m>0. Now, if gradient of AC is 3m, how is this similar to the line AB?
  4. Mar 20, 2010 #3
    Hey thanks for the reply!

    Alright I got h = 4 and -3 and k = 5 and -2, not sure what that means but thank god I don't have to at my level.

    That's what I don't know either, I just wrote the question as is. I remember my friend trying to explain it to me, I think he had Q on the y-axis then the P at some point we were trying to find... or the other way around

    There is a diagram in the book which I couldn't put on here for obvious reasons, but I worked out that the m gradient is 1/2 thanks to an answer I got on yahoo answers, but most his working was too messy to make sense

  5. Mar 20, 2010 #4


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    No problem! :smile:

    Well, the question says find the point (h,k). This means you have two points (4,5) and (-3,-2) which satisfy that they both lie on the line y=x+1 AND they are both a distance of 5 units from the point (0,2).

    Then it's up for debate whether Q is on the x or y axis. If in doubt, why not try both? :tongue:
    P is not what we're trying to find. Lets take the first case where Q is on the x-axis. Then P is on the y-axis. Now we're looking for R which lies on a line with gradient 1/2 passing through Q. Use the point-gradient formula to find the equation of this line. We are told that R has an x-value of 2a, so can you find the y-value?

    Yes, he does make it look confusing :zzz:

    To find m, since you are told that the gradient of AB is -3m, and you can find the actual gradient because you know the points of both A and B, finding the gradient will give you -3/2 so then -3m = -3/2 and m = 1/2.
    Now you know gradient of BC is m and AC is 3m. You know the point of B and A so you'll have two equations in two unknowns that you can solve simultaneously to find the point of intersection between these two lines (the point C).
  6. Mar 21, 2010 #5
    So just to make clear, BC and AC are the two simultaneous equations I have to make?

    I'm also not sure how the guy got (Distance AC) = (2 sqrt(4+9))
    Last edited: Mar 21, 2010
  7. Mar 21, 2010 #6


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    You know that the lines make a triangle, so all you need is a line passing through A and another passing through B and they will connect at C. Solve A and B simultaneously, and you'll get C as your solution.

    And I don't know how that guy got that either :yuck:
  8. Mar 21, 2010 #7
    Ok sorry for dragging this out, but I don't understand what simultaneous equation I should be solving or how to write it out... my attempt was what I wrote above which distance of AC and BC which didn't workout, and equation line using y-y1=m(x-x1) which needless to say didn't workout either, but I thought that's the only way to incorporate the gradient.

    Btw, is it possible to workout out a set of 3 coordinates? in my book for a quadrilateral it goes something like A (1,2) B (3,4,5)?
    Last edited: Mar 21, 2010
  9. Mar 22, 2010 #8


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    We'll drag it out as long as we need to for your to understand it :smile:

    Let's start again.
    1. What is the gradient of the line passing through the points A(1,1) and B(-1,4)?
    2. Since you are told the gradient is also -3m, what does this make the value of m?
    3. The line AC has a gradient of 3m, so what is its gradient using your value of m?
    4. What is the equation of this line, using [itex]y-y_1=m(x-x_1)[/itex], you have your gradient from 3) and you have the x and y coordinates since it's going through the known point A.
    5. Repeat 3 and 4 for the line BC.
    6. Now you have two line equations. They have unknown values x and y. If you solve these two equations simultaneously, this will give you the point of intersection between the two lines AC and BC. This is the point C.
  10. Mar 22, 2010 #9
    Haha turns out that's what I originally did... and turns out my problem all along was making a terribly careless arithmetic mistake :rofl:

    Thanks anyway :smile:

    So what does it mean when a coordinate has A (1,2) and B (3,4,5)?

    And about question 3, would it be a good idea to rewrite the question word-for-word?
    Last edited: Mar 22, 2010
  11. Mar 22, 2010 #10
    Here is the question word-for-word:

    P and Q are the points of intersection of the line y/2 + x/3 = 1 with the x and y axes respectively. The gradient of QR is 1/2, where R is the point with x-coordinate 2a, a>0.

    a. Find the y coordinate of R in terms of a

    b. Find the value of a if the gradient of PR is -2
  12. Mar 23, 2010 #11


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    Haha no problem :biggrin:

    The first defines a point in 2-d space while the second is a point in 3-d space. You won't be dealing with the second for a while, so don't worry about it :smile: and by that time you'll already be really efficient in coordinate geometry (well, you should be anyway) so 3-d is a simple extension on 2-d concepts.

    Oh oops I now realize that R was connected to Q all along! I misread the question hehehe...
    Notice that is says the gradient of QR, which means Q connected to R.

    Ok so it is easy to find what the point of Q is (you already did this in your first post) and since it says QR has a gradient of 1/2, we can now give the equation for the line QR with the point-gradient formula [itex]y-y_1=m(x-x_1)[/itex]. Since R has its x-coordinate at 2a, what is R's y-value then?

    For b) notice that you can again create an equation for the line PR since you're given the coordinates of P and the gradient of the line. Now you have two equations with two unknowns. Since you can find the point R from the intersection of these two lines (by solving them simultaneously) and it says that the x-coordinate of R is 2a, it's easy to find a.
  13. Mar 23, 2010 #12
    Lol turns out the question was (1.2, 3) not (1,2,3) :redface:

    I'm a bit confused here... I found the point of PQ, which one is Q exactly?

    I somehow get y-y= 1/2(x-2a)? but that doesn't work at all
  14. Mar 24, 2010 #13


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    Then ignore everything I said :biggrin:

    Drawing a diagram will be very helpful. Do this for every coordinate geometry question! It doesn't have to be perfectly accurate, just as long as you approximate the line's intersections with the x and y axis, it will give you a good idea of what the graph actually looks like.

    Read the question again, "P and Q are points of intersection [on the x and y axis] at y/2 + x/3 = 1 which becomes (0, 2), (3, 0)."
    So Q is the point (3,0) from what we can best deduce from the question.

    The formula is [tex]y-y_1=m(x-x_1)[/tex] which is the equation of the line that passes through the point [tex](x_1,y_1)[/tex] and has gradient m.

    The equation of line QR is [tex]y-0=\frac{1}{2}(x-3)[/tex] since Q is the point (3,0) and QR has gradient 1/2.

    Now, this equation of the line basically says that if there is a point (x,y) that lies on this line, substituting the values of x and y into the equation will give you 0=0 (or in other words, the left hand side of the equation will equal the right hand side). If it doesn't lie on this line, you won't get 0=0

    e.g. lets try the point (1,-1) to see if it lies on the line [tex]y=\frac{1}{2}(x-3)[/tex]
    Substitute x=1 and y=-1 into the equation:
    Therefore (1,-1) lies on this line.

    Trying, say, the point (1,0) will not give you this result, it actually gives you [itex]0=-1[/itex] so this tells you that point doesn't lie on the line.

    In a similar way, we can go backwards. If we want to find what y-value the point with x=3 has, we simply substitute x=3 into the line equation and solve for y:
    Therefore the point (3,0) lies on the line.
    This can also be done for finding the x-value if we know the y-value of a point on a line.

    Ok back to the question:
    We have been told that R has an x-coordinate 2a, so can you find the y-value of R now? :smile:
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