coordinate

niharika_kumar · 2014-09-10 04:15:54

If the vertices of a triangle are

, then find its orthocentre.

Bob · 2014-09-10 06:20:54

hi Niharika,

(1) Get an expression for the gradient of one side.

(2) The gradient of the altitude to that side will be minus the reciprocal of that answer.

(3) Hence get the equation for one altitude in the form y = mx + c.

(4) Use the symmetry of the coordinates to write down the equation of a second altitude.

(5) Subtract to find the x coordinate where they intersect.

You'll find there is some simplification at each step and the final result is quite simple.

Bob

anonimnystefy · 2014-09-10 06:25:59

Not sure if this is correct, but I am getting:

Last edited by anonimnystefy (2014-09-10 06:31:09)

Bob · 2014-09-10 06:28:09

That's what I have. Please hide your answer so N has a chance to do it before seeing yours.

Bob

anonimnystefy · 2014-09-10 06:32:18

Yes, sorry. I was just checking to see if I have messed up.

I need to get into form a bit, haven't done much math in the last 3 months.

Agnishom · 2014-09-10 12:49:22

Hi bob, I am stuck at step 2

anonimnystefy · 2014-09-10 12:54:04

What are you stuck with?

Agnishom · 2014-09-10 19:42:36

I cannot parse the sentence. Can you rephrase it?

Bob · 2014-09-10 19:50:51

hi Agnishom,

If one line has a gradient m, then a perpendicular line will have gradient -1/m.

Bob

Agnishom · 2014-09-10 20:51:12

Yes, that is true.

anonimnystefy · 2014-09-11 01:17:50

Did you get the answer?

niharika_kumar · 2014-09-11 01:22:19

are you sure x=a and not -a.
I am getting x=-a.

anonimnystefy · 2014-09-11 01:26:46

Hm, what are your altitude equation?

Bob · 2014-09-11 03:49:42

yes, x = -a is good.

And the expression for y is symmetrical in t; by which I mean if you swap over t1 and t2 (or t3) the answer is unchanged.

Bob

niharika_kumar · 2014-09-11 04:11:51

thanks bob.

niharika_kumar · 2014-10-01 19:11:03

Here is another question:-

Obtain the condition that one of the straight lines given by the equation

may coincide with one of those given by the equation
a'x^2+2h'xy+b'y^2=0

Bob · 2014-10-01 20:21:03

hi Niharika,

I'm not at all sure about this at the moment so I'll work on it. Here's what I know:

In the theory of conic sections, you start with a cone (and an inverted identical cone sitting so their vertices coincide) and then take slices through to generate a circle, ellipse, parabola or hyperbola. Some sources also recognise a 5th section: if the slice cuts through the edges of the cone then the section is a pair of straight lines. Some sources call this a degenerate case.

The equation for this is:

Each bracket gives one of the line equations.

As x = y = 0 is a point of your system the equation in this case must simplify to

and if a second system has one line in common, it must have the form

So we end up with:

That's as far as I've got for now. Hopefully back later.

Bob

Bob · 2014-10-02 00:26:20

Ok. Some progress. The above has too many variables for my liking so I thought I'd reduce it by writing the lines in the form y = mx.

As I haven't found a way to do 'a-dash' in Latex, I'll use capitals for a', b' and h' ... ie. A, B and H.

So

This leads to

and similarly

You can eliminate m from these by dividing the first by the third and subtracting the fourth from the second:

From which you can get m, n and p each in terms of a, A, b, B, h and H.

These are very similar expressions, so I suspect you can do something to show perhaps m^2 = np let's say. I haven't tried this yet so that might not be it.

Bob

Bob · 2014-10-02 00:38:53

Actually, I've just got

I think that will do.

Bob

niharika_kumar · 2014-10-02 16:16:34

thanks bob.

I didn't consider fact that y-mx is the factor of both the equations, but yes you used it, so
if we have that (y-mx) is the factor of both the given equations,
why don't we put y=mx in both of them, then eliminate the term x^2 and solve by cross multiplication.

that gives m^2 =(hA-Ha)/(bH-hB)={(aB-Ab)/2(bH-Bh)}^2

That has given a clear solution without m,n and p and the steps are less complicated.

Last edited by niharika_kumar (2014-10-02 16:17:10)

Bob · 2014-10-02 19:36:43

hi Niharika,

Your suggestion looks good to me. Well done for finding a quicker way.

More good news: I can now do this

My thanks to Olinguito for this.

Bob

niharika_kumar · 2014-10-02 22:34:55

my thanks to him too.

Even I was unable to do that and hence my 2nd equation was not in latex form.

niharika_kumar · 2014-10-03 16:23:26

here is one more:-

Show that all the chords of the curve

which subtend a right angle at the origin pass through a fixed point.Also find that point.

Bob · 2014-10-04 08:39:49

hi Niharika,

I've tried this, on and off, all day and I cannot come up with a neat solution. The best I can offer has horrible algebra:

(1) Call the lines y = mx and y = -x/m

(2) Get the coordinates, in terms of m, where these intersect the curve.

(3) Hence the equation of a chord in terms of m.

(4) Find an 'm' independent point on this line.

Bob

Math Is Fun Forum

#1 2014-09-10 04:15:54

coordinate

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Re: coordinate

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Re: coordinate

#5 2014-09-10 06:32:18

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