geometry help!

J

jennsunn

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Joined
Oct 7, 2005
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17
I need help with these three questions:

1. The graphs of the lines 3x + 4y = 2, 5x - y =11, and x +ky = -2, all contain the same point. Find the point and also the value of k.

2. A circle has its center at the point (-2,3). If one endpoint of the diameter is at (3,-5), find the other endpoint.

3. There are two points on the x - axis at a distance of 6 units from the point (3,1). Determine the coordinates of each point.

Thank you so much!
 
1) Use the first two equations to find the mutual intersection point. Then plug the x- and y-values from this point into the third equation, and solve for k.

2) Hint: What is the relation between a circle's center and the segment which is its diameter?

3) Plug (x, 0) and (3, 1) into the Distance Formula, insert the desired distance, and solve for x.

Eliz.
 
G'day Jennsunn,

The graphs of the lines 3x + 4y = 2, 5x - y =11, and x +ky = -2, all contain the same point. Find the point and also the value of k.
Click to expand...
If the three lines intersect at a common point, then the first two lines will intersect at that point as well.

So you can solve the simultaneous equations
3x + 4y = 2 and
5x - y =11

To find the point of intersection of these two lines and hence the intersection of all three lines.

The line given by x +ky = -2 must pass through that point. So if those two lines intersected at, say, (2,-1) then you can substitute in x=2 and y=-1 into x +ky = -2 to find k.
 
Nice work.

For the second one, we have:
Code: 
            y /|\
               |
               |
               |
               |
centre -> *   3+
     /|\       | 
      |        |     
  ----|---+----+-----+-------->
      |  -2    |     3        y
      |        |
 up   |        |
 8    |        |
      |        + -5  *  <- one endpoint
     \|/       |            of the diameter
          <---------->
            across 5
So the other end point must be 5 to the left and 8 up from the centre, agreed?
 
For the third one.

If you go with Eliz's approach then you have
\(\displaystyle 6 = \sqrt{\, (x - 3)^2 + (0 - 1)^2 \, }\)

Simplify what's under the radical. Square both sides of the equation to remove the radical.

Solve the resulting quadratic for x. The coordinates you desire have these x-ordinates, and are on the x-axis so have y-ordinates of 0.

Edit: just saw your latest post now.
 
Re: help with answer

jennsunn said:
I still do not get it...
howd you get so smart? ughhh!
Click to expand...
The length between the centre of a circle and one of the end points of its diameter is the radius, right?

So to find the other end point of the diameter, you would go "backwards" that same length, right?


By the way, I've never been and will never be smart. This is only experience.
 
ok, dont think that i am dumb but would it be -3,2? By the way..you could have fooled me, i think it does have to do with smarts
 
I would never think that.

This is what the diameter does:
Code: 
*
            y /|\
   *           |
               |
      *        |
               |
centre-> *    3+
               | 
            *  |     
  -------+-----+-----+-------->
        -2     *     3        y
               |  
 up            |  *
 8             |       
               + -5  *  <- one endpoint
               |            of the diameter
          
   (The scaling looks horrible, I know... but 
     hopefully you get the idea.)
Hopefully the moving left 5 and up 8 from the centre from the previous diagram makes a little more sense.
 
second try at an answer

ok second attempt for # 2...
Am I right by this?
for x:
(1x + 2x) /2 = -2
(3x + x)/ 2 = -2
3 + x = -4
x = -7
same for y:
(y1 + y2)/2 = 3
(-5 +y)/ 2 = 3
-5 + y = 6
y = 11
so the answer is (-7,11)
 
Re: second try at an answer

jennsunn said:
ok second attempt for # 2...
Am I right by this?
for x:
(x1 + x2) /2 = -2
(3 + x)/ 2 = -2
3 + x = -4
x = -7
same for y:
(y1 + y2)/2 = 3
(-5 +y)/ 2 = 3
-5 + y = 6
y = 11
so the answer is (-7,11)
Click to expand...
Excellent work!!!
 
ok..this is what i tried let me knows how i did

1^2 + b^2 = 6^2

1 + b^2 = 36

b^2 = 35

b = radical 35

So our coordinates would be (3 + radical 35, 0) and (3 - radical 35, 0)
 
Ok.

It doesn't feel right to be to have two points (x,0), so I'm going to let \(\displaystyle (x_0, 0)\) and \(\displaystyle (x_1, 0)\) be the two points on the x-axis which are 6 units away from the point (3,1). \(\displaystyle (x_0, 0)\) will be on the left, \(\displaystyle (x_1, 0)\) will be to the right.

The distance between two points \(\displaystyle (x_a , y_a)\) and \(\displaystyle (x_b, y_b)\) is given by
  • \(\displaystyle D = \sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}\)

I think you are ok with that.

Now if we first consider the distance between the point \(\displaystyle (x_0, 0)\) and (3,1):

We have a distance of 6, D = 6. The two points to plug in are \(\displaystyle (x_0, 0)\) and \(\displaystyle (3, 1)\). So replace \(\displaystyle (x_a , y_a)\) and \(\displaystyle (x_b, y_b)\) with these, respectively.

  • \(\displaystyle 6 = \sqrt{(x_0 - 3)^2 + (0 - 1)^2}\)

Simplify what's under the radical:

  • \(\displaystyle 6 = \sqrt{(x_0 - 3)^2 + 1}\)

Square both sides to remove the radical

  • \(\displaystyle 36 = (x_0 - 3)^2 + 1\)

Move that 1 over to the left-hand side:

  • \(\displaystyle 35 = (x_0 - 3)^2\)

Take square roots and solve for \(\displaystyle x_0\):

  • \(\displaystyle x_0 = \pm \sqrt{35} + 3\)

Take the negative square root of 35 for \(\displaystyle x_0\) because it's the one on the left:

  • \(\displaystyle x_0 = -\sqrt{35} + 3\)

\(\displaystyle x_1\) will come from taking the postive square root of 35:

  • \(\displaystyle x_1 = \sqrt{35} + 3\)

And the coordinates follow.

Edit: I've just seen your work. Very smooooooothhhh...............
 
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