Well... I'm stumped. (straight line graps) (1 Viewer)

[FDownes]

Well... I'm stumped. (straight line graphs)

I'm working through Challenge Exercise 7 of Maths in Focus 1, and although I'm only on the first question, I'm already stumped! I'll only post this one here for now, but these challenge exercises always seem to be a lot of trouble, so I'll probably have more later on.

Question 1:

If points (-3k, 1), (k - 1, k - 3) and (k - 4, k - 5) are collinear, find the value of k.

Collinear means they all lay on the same line. This also means they will all have the same gradient. This much I know. I'll keep working on it, and update this thread with edits/replies if I get anywhere.

SOLVED IT!

= (k - 3 - 1) / (k - 1 + 3k) = (k - 5 - 1) / (k - 4 + 3k) [since the gradients between the points should be equal, just do two different forms of the gradient formula]
= (k - 4)(4k - 4) = (k - 6)(4k - 1)
= 4k^2 - 20k + 16 = 4k^2 - 25k + 6
= 5k = -10
= k = -2

Simple.

 

[FDownes]

Uh oh... Got another problem that I can't solve.

Find the equation of the circle whose center is at the origin and with tangent x - 3y + 9 = 0.

The perpendicular of the tangent of a circle will always pass through the origin, if taken from the point where it touches the circle. The gradients of two perpendicular lines will multiply to give -1. The equation of a circle with its origin at the center will be x^2 + y^2 = r^2. I know this much, and although I think I know how I should be going about this, I can't seem to get the correct answer. Please help!

 

[joey_prince42]

Uh oh... Got another problem that I can't solve.

Find the equation of the circle whose center is at the origin and with tangent x - 3y + 9 = 0.

The perpendicular of the tangent of a circle will always pass through the origin, if taken from the point where it touches the circle. The gradients of two perpendicular lines will multiply to give -1. The equation of a circle with its origin at the center will be x^2 + y^2 = r^2. I know this much, and although I think I know how I should be going about this, I can't seem to get the correct answer. Please help!

Hey there. The question is easy - all you need to do is work out the radius of the circle and since the tangent to the circle is perpendicular to the circle, you can use the perpendicular distance formula from the point (0,0) which gives you 9/sqrt(10) so the equation of the circle must be x^2+y^2=81/10 .

Hope that answers your question,

Joey

 

[FDownes]

Eep! I completely forgot I posted this here! Sorry!

As for the question, I solved it a short time after posting it here (I had forgotten about the perpendicular distance formula), but thanks for your help anyway.

 

[FDownes]

Got another problem I need help with...

Find the coordinates of the center of a circle that passes through the points (7,2), (2,3) and (-4,-1)

Can anyone help me out here?

 

[crazysambo]

couldn't you just find teh distance between two of teh points and halve it. im not thinking straight, im in dt at the moment, ill try it wen i get home.

 

[Mark576]

umm ...that would only work if two co-ordinates were given as the length of the diameter, which they aren't.
Try using the equation of a circle ...then sub in a few points, solve simultaneously

 

[crazysambo]

yer i got it out. it took me a bit of writing. wat i did was first start with (7-x)^2 + (2-y)^2 = (2-x)^2 + (3-y)^2, then u equate them and u end up with y= something, sori i dont have my book. so thts ur first equation. then u do (7-x)^2 + (2-y)^2 = (-4-x)^2 + (-1-y)^2, and u equate that and u get a second eqation, then u solve the 2 equations simultaeneously and im pretty sure form memory x = 3 and y =2 so then ur point is (3,2) i think thts wat i got.