Maths help (1 Viewer)

[D94]

Recognise that you'll need to use the expression for a circle:

(x - h)² + (y - k)² = r²

where (h,k) is the centre.

Substituting in the centre, and then substituting in the given point, we get the radius to equal sqrt(101). Multiplying that by two, we get 2*sqrt(101) as the diameter.

 

[planino]

Well if you think about it, any point on the circumference of the circle is always the same distance away from the centre, right? If you find the distance between that point which the circle passes through and the centre, you get the radius, which is half the diameter.




um the question is just asking for the diameter, which is two times the radius. You don't need the circumference formula, there is no pi in the solution

AHH CRAP I feel like an idiot now

 

[twinklegal19]

Aren't I finding the diameter? I won't need the circumference formula. I'll do:
View attachment 25635
View attachment 25636

Is that right?

yeah that looks right

AHH CRAP I feel like an idiot now

haha that's okay

 

[Shazer2]

Alright cool. I'm moving onto a midpoint exercise now, so I'll post if I have some troubles.

 

[Shazer2]

Alright, so the question is, The points A(3, 5), B(9, -3), C(5, -6) and D(-1, 2) form a quadrilateral. Prove that the diagonals are equal and bisect one another. What type of quadrilateral is ABCD?

I got the midpoint for AC and BD, and that is (4, -0.5) but in the back before that it has AC = BD = sqrt(125). I'm not sure what that part is, and they also state it's a rectangle, I'm not sure how they find that either. Help please!

 

[RealiseNothing]

Alright, so the question is, The points A(3, 5), B(9, -3), C(5, -6) and D(-1, 2) form a quadrilateral. Prove that the diagonals are equal and bisect one another. What type of quadrilateral is ABCD?

I got the midpoint for AC and BD, and that is (4, -0.5) but in the back before that it has AC = BD = sqrt(125). I'm not sure what that part is, and they also state it's a rectangle, I'm not sure how they find that either. Help please!

That should be 11.1803 (I had to). But the mid point isn't equal to the length of AC or BD.

 

[Shazer2]

It has AC = BD = sqrt(125). Midpoint AC = Midpoint BD = (4, -0.5); rectangle as the answer.

 

[twinklegal19]

Alright, so the question is, The points A(3, 5), B(9, -3), C(5, -6) and D(-1, 2) form a quadrilateral. Prove that the diagonals are equal and bisect one another. What type of quadrilateral is ABCD?

I got the midpoint for AC and BD, and that is (4, -0.5) but in the back before that it has AC = BD = sqrt(125). I'm not sure what that part is, and they also state it's a rectangle, I'm not sure how they find that either. Help please!

Well you got the first half of it, you already proved it to be a parallelogram. However you need to be more specific by finding the length of the diagonals as well, which in this case are equal. A rectangle has equal diagonals that bisect each other.

That should be 11.1803 (I had to). But the mid point isn't equal to the length of AC or BD.

m8

 

[Shazer2]

Alright thanks If I have the radius of a circle, and the midpoint is given, how can I find the coordinate of the missing part of the line (to make up the diameter). SO the centre is (-2, 5) and the end of the diameter is (4, -3). Find the coordinates of the other end of the diameter. The information I am given is the midpoint and the point needed to get the distance from the centre.

 

[RealiseNothing]

m8

What?

Did you expect me not to?

 

[deswa1]

Alright thanks If I have the radius of a circle, and the midpoint is given, how can I find the coordinate of the missing part of the line (to make up the diameter). SO the centre is (-2, 5) and the end of the diameter is (4, -3). Find the coordinates of the other end of the diameter. The information I am given is the midpoint and the point needed to get the distance from the centre.

The simple way to do this is as follows:
Note that to go from the centre (-2,5) to one end of the radius (4,-3) means that you go 'up' by (6,-8). The other end of the circle will be the exact same distance just in the opposite direction so go 'down' by (6,-8). Therefore, the other end is at the point (-8,13)

 

[Shazer2]

Is there an actual way to work it out? I'm going to fail this test tomorrow hardcore. Why do I suck at maths so much!

 

[deswa1]

Is there an actual way to work it out? I'm going to fail this test tomorrow hardcore. Why do I suck at maths so much!

Yes, you could use the two points to work out the equation of the line and then use the distance formula to work out what point on that line lies the distance of the radius away. For these problems though, if you draw a diagram and have a decent conceptual understanding it's much easier.

Using that question as an example, the centre of a circle has to be in the middle. Therefore, if the x coordinate of one end of the circle is Z units away in one direction, the other end of the circle has to be Z units away in the other direction. Draw a diagram to try and visualise what I'm saying

 

[twinklegal19]

What?

Did you expect me not to?

No, of course not

it's the most obvious thing in the world [that you "had" to do it]

Is there an actual way to work it out? I'm going to fail this test tomorrow hardcore. Why do I suck at maths so much!

well a straightforward way is to use the midpoint formula, which you sub in the point on the circle and the centre.

 

[Shazer2]

If I sub the center coordinate in and the coordinate lying on the circle, the midpoint I get is (1, 1). Am I going to fail 2U maths if I can't do basic things?

 

[deswa1]

If I sub the center coordinate in and the coordinate lying on the circle, the midpoint I get is (1, 1). Am I going to fail 2U maths if I can't do basic things?

You aren't trying to find the midpoint of the center and the edge of the circle. The point you are trying to find is P(x,y). The midpoint of P and the edge of the circle is the centre of the circle. In this case, you aren't finding the midpoint but using the midpoint that you are given to find one of the other points

 

[twinklegal19]

If I sub the center coordinate in and the coordinate lying on the circle, the midpoint I get is (1, 1). Am I going to fail 2U maths if I can't do basic things?

You already have your midpoint, which is the circle. You sub in the centre and point lying on the circle to get the other point (on the other end of the diameter) and rearrange.

Well, are you just weak at this topic or in general? If I recall correctly, you should have learnt all of these in year 8/9

The best thing you can do right now is to catch up with the basics as soon as possible and ask your teacher for help for anything.

 

[Shazer2]

I've always got this done in class, then I come home and I'm stumped. I'm one of the better students in the class overall.

I can't get this, and it's annoying me. :'( Mind if I PM you?

 

[twinklegal19]

I've always got this done in class, then I come home and I'm stumped. I'm one of the better students in the class overall.

I can't get this, and it's annoying me. :'( Mind if I PM you?

Sure

 

[Shazer2]

Alright, another problem.

I need to find the equation of a straight line passing through (0, -2) and perpendicular to the line x - 2y = 9. What I have tried to do so far is get the gradient of the line which is 1/2 and from there, I'm not sure what to do.

Help please