# A couple of Questions.

• Apr 8th 2009, 12:00 AM
Blahdkm
A couple of Questions.
I have to find all points of intersection of the circle "x^2 + (y-1)^2=25" and the line "y=-7x+26" using a system of equations. How?

I also have the triangle with Vertices at A(13,13), B(9,3), and C(-1,-1). From that I have to find the equation of the median AM(M is the midpoint of BC), and the equation of BN(N is the midpoint of AC), what exactly is the "equation of the median".

Secondly, I have to use these equations to find P(the point of intersection of AM and BN), how?

Then I must prove the distance between P and M is 1/3 the length of the median AM, why?

All help is much appreciated.
• Apr 8th 2009, 01:36 AM
running-gag
Quote:

Originally Posted by Blahdkm
I have to find all points of intersection of the circle "x^2 + (y-1)^2=25" and the line "y=-7x+26" using a system of equations. How?

I also have the triangle with Vertices at A(13,13), B(9,3), and C(-1,-1). From that I have to find the equation of the median AM(M is the midpoint of BC), and the equation of BN(N is the midpoint of AC), what exactly is the "equation of the median".

Secondly, I have to use these equations to find P(the point of intersection of AM and BN), how?

Then I must prove the distance between P and M is 1/3 the length of the median AM, why?

All help is much appreciated.

Wow ! So many questions !

M(x,y) is on the circle iff $\displaystyle x^2 + (y-1)^2=25$
M is on the line iff $\displaystyle y=-7x+26$

M is on the circle and on the line iff (x,y) satisfy the system
$\displaystyle x^2 + (y-1)^2=25$
$\displaystyle y=-7x+26$

To solve you just need to substitute y in the first equation using the second equation. You will get a quadratic equation (unknown x) which can have 0 (no intersection point), 1 (1 intersection point = the line and the circle are tangent) or 2 solutions depending by the value of the discriminant.

As soon as you have found the values for x, you substitute in the second equation to get the values for y.

A(13,13), B(9,3), and C(-1,-1).
M being the midpoint of (BC) : $\displaystyle x_M = \frac{x_B + x_C}{2} = 4$ and $\displaystyle y_M = \frac{y_B + y_C}{2} = 1$
(AM) passes through A(13,13) and M(4,1). You should be able to find its equation now.

As soon as you have the equation of (AM) and (BN) you can solve the system to get the coordinates of P.
• Apr 8th 2009, 01:48 AM
earboth
Quote:

Originally Posted by Blahdkm
(1) I have to find all points of intersection of the circle "x^2 + (y-1)^2=25" and the line "y=-7x+26" using a system of equations. How?

(2) I also have the triangle with Vertices at A(13,13), B(9,3), and C(-1,-1). From that I have to find the equation of the median AM(M is the midpoint of BC), and the equation of BN(N is the midpoint of AC), what exactly is the "equation of the median".

Secondly, I have to use these equations to find P(the point of intersection of AM and BN), how?

Then I must prove the distance between P and M is 1/3 the length of the median AM, why?

All help is much appreciated.

to (1):

You already have a system of 2 equations:

$\displaystyle \left|\begin{array}{rcl}x^2+(y-1)^2&=&25\\ y&=&-7x+26\end{array}\right.$

Plug in the term of y of the 2nd equation into the 1rst equation:

$\displaystyle x^2+(-7x+26-1)^2=25$

and solve this equation for x. You'll get the x-coordinates of the points of intersection. To calculate the y-coordinates of the intersection points plug in the values of x into the equation of the line.

I've got $\displaystyle P_1(3, 5)~,~P_2(4, -2)$

to (2)

Calculate the coordinates of M: $\displaystyle M\left(\dfrac{9+(-1)}{2}~,~\dfrac{3+(-1)}2\right)$

Now use the 2-point-formula of a straight line

$\displaystyle \dfrac{y-13}{x-13}=\dfrac{4-13}{1-13}= \dfrac34$

That means: $\displaystyle AM: y = \dfrac34 x+\dfrac{13}4$

Calculate the equation of BN in exactly the same way.

The point of intersection between AM and BN is called the centroid (google for it) and it divides the medians in the relation of 2 : 1 from the vertex.
• Apr 8th 2009, 05:01 AM
Blahdkm
Quote:

Originally Posted by earboth
Now use the 2-point-formula of a straight line

$\displaystyle \dfrac{y-13}{x-13}=\dfrac{4-13}{1-13}= \dfrac34$

That means: $\displaystyle AM: y = \dfrac34 x+\dfrac{13}4$

Calculate the equation of BN in exactly the same way.

The point of intersection between AM and BN is called the centroid (google for it) and it divides the medians in the relation of 2 : 1 from the vertex.

Many thanks to both of you, just one question, where did the 13/4 come from?

Also, did you mix up the x and y coordinate on the second equation? I got 4/3 my first time through.(Mostly self-check right now)

Also once more, how do you calculate the centroid?
• Apr 8th 2009, 05:32 AM
masters
Quote:

Originally Posted by Blahdkm
Many thanks to both of you, just one question, where did the 13/4 come from?

Also, did you mix up the x and y coordinate on the second equation? I got 4/3 my first time through.(Mostly self-check right now)

Also once more, how do you calculate the centroid?

Hi Blahdkm,

The midpoint M of BC is (4, 1)

The slope of AM is $\displaystyle \frac{1-13}{4-13}=\frac{4}{3}$

Using $\displaystyle y=mx+b$ with $\displaystyle m=\frac{4}{3}$ and M(4, 1),

$\displaystyle 1=\frac{4}{3}(4)+b$

$\displaystyle b=-\frac{13}{3}$

The equation for AM is $\displaystyle y=\frac{4}{3}x-\frac{13}{3}$

Now, find the equation for BN the same way. Use the two equations to solve for the intersection which will be your centroid.
• Apr 8th 2009, 09:54 AM
earboth
Quote:

Originally Posted by Blahdkm
...

Also, did you mix up the x and y coordinate on the second equation? I got 4/3 my first time through.(Mostly self-check right now)

Also once more, how do you calculate the centroid?

You are right and I'm sorry for the confusion.

To check your final result use the following formula:

If the vertices of a triangle are $\displaystyle A(a_1, a_2)\ ,~B(b_1, b_2)\ ,~C(c_1, c_2)$then the centroid P has the coordinates:

$\displaystyle P\left(\frac13(a_1+b_1+c_1)\ ,~\frac13(a_2+b_2+c_2)\right)$

With your vertices you'll get P(7, 5)