Math Is Fun Forum

1a2b3c2212

Member

Registered: 2009-04-04

Posts: 419

Co-ordinate Geometry

Three points A(6,10), B(-2,4) and C(h,-4) form a triangle ABC.
Midpoint of AB is M. The line through M meets AC at the point such that AP:PC=1:3

a)Given that AB=BC and that h is positive, find h

b) Find coordinates of M and P

c) find area of the quadrilateral BMPC.

For b),How to find the coordinates of P and what's the fastest way?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Co-ordinate Geometry

Hi 1a2b3c2212;

For a)

I am getting h = 4:

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

1a2b3c2212

Member

Registered: 2009-04-04

Posts: 419

Re: Co-ordinate Geometry

uh ya.=)

how to do part b)?? there's many ways of solving it but any idea of which is the simplest to understand?

careless25

Real Member

Registered: 2008-07-24

Posts: 560

Re: Co-ordinate Geometry

To find the mid point of AB. You add x co-ordinates of A and B and divide by 2. This gives you the x co-ordinate of the midpoint M. then add the y co-ordinate of A and B and divide by 2.

M should be (2,7).

For P: I m not sure but this should work. Add the x coordinate of A and C divide by 3. this should be x coordinate of P and then repeat for y co-ordinate.
i get (10/3,2) for P.

Lets see what bobbym says. I know M is correct but i m not sure about P.

C25

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Co-ordinate Geometry

Hi guys;

Lets see what bobbym says.

I asked him and he said (2,7) is correct for M. But he didn't agree with P, because (10/3 , 2) is not on the line AC.

Cautiously, I like (11 / 2 , 13 / 2) for P

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

careless25

Real Member

Registered: 2008-07-24

Posts: 560

Re: Co-ordinate Geometry

Thanks bobbym!

1a2b3c2212

Member

Registered: 2009-04-04

Posts: 419

Re: Co-ordinate Geometry

Hi bobbym,

P(5.5,6.5) is the correct answer. How did you arrive at it?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Co-ordinate Geometry

Hi 1a2b3c2212;

Distance of the line segment AC = 10 sqrt(2).Since the segments AP and PC are 1:3 then the distance of AP = 10 sqrt(2) / 4

Now I found the eqtn of the line AC to be y = 7x - 32. So we need some point (x,y) on AC that has length 10 sqrt(2) / 4 and lies on y = 7x - 32.

Using y = 7x-32 we can eliminate the y:

x = { 11 / 2 , 13 / 2}

Plugging them both back into y = 7x-32 you can see that x = 13 / 2 is nonsensical : So

x = 11 / 2 , y = 13 / 2 are the coordinates of P

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

1a2b3c2212

Member

Registered: 2009-04-04

Posts: 419

Re: Co-ordinate Geometry

why is 13/2 nonsensical?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Co-ordinate Geometry

Hi 1a2b3c2212;

When you plug it into y = 7x - 32 the equation that the line segment AC rests on, you get 13 / 2 which is not on AB. In other words it is not between (6,10) and (4,-4),

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.