Wednesday

April 23, 2014

April 23, 2014

Posted by **Anonymous** on Wednesday, February 27, 2008 at 11:08am.

1. Consider the triangle ABC, with vertices A(0,-2), B(9,1), C(-1,11).

Find:

a) a cartesian equation of the altitude from C;

b) a cartesian equation of the circle which has BC as a diameter;

c) a cartesian equation of the circle with C and the line through A and B as a tangent

d) the points of intersection of the circles described in b) and c)

?*cartesian equation is y=mx+c

Please help. I really don't get this.

- Math -
**drwls**, Wednesday, February 27, 2008 at 11:41amNot all Cartesian equations are straight lines. It is any equation that uses the variables x and y. For example, b) and c) are circles.

For (a), they want the equation of a line through C that is perpendicular to the side c (from A to B)

The line from A to B has slope

m = (1+2)/(9-0) = 1/3

Therefore the altitude has slope -3.

The equation for the altitude from C is

(y-11)= -3(x+1)

y = -3x + 8

b) BC diameter has a center at (4,6). and that is the center of the circuscribed circle . The length of the diameter is sqrt[(10^2 + 10^2)]

= sqrt 100 = 10 sqrt 10. The radius of the circle around it is 5 sqrt10

From that information, the equation of the circle is

(x-4)^2 + (y-6)^2 = [5 sqrt10]^2 = 50

c) I think you mean "circle with center at C" and line through A and B as tangent. Get the equation of that line and determine its distance from C.

You are going to have to take it from here. It's time for you to practice what you've learned. Someone will be glad to critique your work

- Math -
**Reiny**, Wednesday, February 27, 2008 at 11:47amDid you draw a diagram?

Most people are "visual" and it will make the question easier to see.

a) the altitude must meet AB and must be perpendicular to AB

so slope of AB = 3/9 = 1/3

which makes the slope of the altitude -3

we know the altitude has slope -3 and must pass through (-1,11), so....

11 = -3(-1) + b, (I am using y = mx+b)

b=8

altitude equation: y = -3x + 8

b) are you familiar with the general equation of the circle:

(x-h)^2 + (y-k)^2 = r^2, where the centre is (h,k) and the radius is r ??

if so, then the centre must be the midpoint of BC which is (4,6)

so equation is

(x-4)^2 +(y-6)^2 = r^2

but (9,1) lies on our circle, so

(9-4)^2+(1-6)^2=r^2

r^2=50

then (x-4)^2 +(y-6)^2 = 50

c)If AB is to be a tangent of the circle with centre at C, then the altitude we found in a) must be the radius of our circle. (Can you see how important a diagram is??)

I hope you have seen the formula to find the distance between a point (p,q) and the line Ax + By + C = 0

it says Dist = │Ap + Bq + C│/(A^2+B^2)

First we need the equation of AB which in general form is x - 3y - 6 = 0

so radius of our circle is

│1(-1) + (-3)(11) - 6│/√(1+9)

= 40/√10

equation of circle:

(x+1)^2 + (y-11)^2 = 40^2/√10^2

(x+1)^2 + (y-11)^2 = 1600/10 = 160

d) I will let you do that one, here is the method

1. expand each of the two circle equations, each will contain an x^2 and a y^2 term

2. subtract one equation from the other, the square terms will drop away, leaving an equation in x and y.

3. solve for either x or y, whichever seems easier

4. substitute that back into the first circle equation

5. You should get two solutions, sub that back into the x and y equation you got in step 2.

Good luck

Let me know if it worked for you

- Math (correction) -
**drwls**, Wednesday, February 27, 2008 at 11:47amThis is what I should have written:

b) BC diameter has a center at (4,6). and that is the center of the circumscribed circle . The length of the diameter is sqrt[(10^2 + 10^2)]

= sqrt 200 = 10 sqrt 2. The radius of the circle around the diameter is 5 sqrt 2

From that information, the equation of the circle is

(x-4)^2 + (y-6)^2 = [5 sqrt2]^2=50

- Math (correction) -
**Reiny**, Wednesday, February 27, 2008 at 11:56amdrwls, I was wondering what a "circuscribed circle" was, lol

not as serious though as several of my students who somehow wanted to circumcise a circle.

BTW, I wish somebody could come up with a way to avoid two or more tutors working on the same problem at the same time.

- To Reiny -
**DrBob222**, Wednesday, February 27, 2008 at 1:02pmThe English/social studies/history et al tutors on this board suggested at the start of the school year that they might avoid two people doing a great deal of work on one question as follows:

For a question involving proofing a paper or something similar in which a lot of work was required, that the first tutor to start on the project post a note that s/he is working on the response and will post the response later. That way, other tutors know the problem is being taken care of and will move on to another subject/question. You might email Writeacher or Ms Sue for more details.

- To Reiny -
- Math (to Reiny et al) -
**drwls**, Wednesday, February 27, 2008 at 1:48pmI'd rather not IM a lot of people to tell them what I'm working on, though doing so can avoid multiple answers and wasted effort. I answer most questions late at night when no one else is around, anyway. I'm quite often wrong with my sloppy math, as you know; so another point of view helps. When the answers are short, not much effort is wasted.

- Math (to Reiny et al) -
**Ms. Sue**, Wednesday, February 27, 2008 at 2:16pmWe in the English/social studies fields sometimes will post a brief message on the board, saying that we're working on a long answer and that we'll post the completed answer in a little while.

- Math (to Reiny & Sue) -
**drwls**, Wednesday, February 27, 2008 at 2:24pmI like that idea

- Math (to Reiny & Sue) -

- Math (to Reiny et al) -

- Math (correction) -

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