Finding the minimum area of triangle, coordinate plane
Hi Forum!
I've been reviewing the things I know so far, and found this question
A line is drawn through (2,1), forming a right triangle at the 1st quadrant. Determine the centroid of the triangle with minimum area.
https://lh3.googleusercontent.com/l...0/Centroid.JPG
Now, if we have the point at (2,1), can the area decrease any further?
Isn't this the smaller area we can possibly obtain?
Why do 2+x and 1+y are necessary?
Thanks!
Re: Finding the minimum area of triangle, coordinate plane
Quote:
Originally Posted by
Zellator Hi Forum!
I've been reviewing the things I know so far, and found this question
A line is drawn through (2,1), forming a right triangle at the 1st quadrant. Determine the centroid of the triangle with minimum area.
https://lh3.googleusercontent.com/l...0/Centroid.JPG
Now, if we have the point at (2,1), can the area decrease any further?
Isn't this the smaller area we can possibly obtain?
Why do 2+x and 1+y are necessary?
Thanks!
point (2,1) slope is m.
Line equation: y=mx2m+1
y=0 ==> x={2m1}/m
x=0 ==> y=12m
S is area of triangle,
S=x*y/2 = {{2m1}/m}{12m}/2
Now find when m so that min [ {{2m1}/m}{12m}/2 ].
When you find this m, you can find x and y, and afterwards the centroid.
Re: Finding the minimum area of triangle, coordinate plane
Quote:
Originally Posted by
Also sprach Zarathustra point (2,1) slope is m.
Line equation: y=mx2m+1
y=0 ==> x={2m1}/m
x=0 ==> y=12m
S is area of triangle,
S=x*y/2 = {{2m1}/m}{12m}/2
Now find when m so that min [ {{2m1}/m}{12m}/2 ].
When you find this m, you can find x and y, and afterwards the centroid.
Hi Also sprach Zarathustra!
Thanks for your reply!
That's great to know, as well.
An alternative way to do it is with 2+x and 1+y, as I previously mentioned.
As it turns out, m=1/2
So if we plug this into your equation we get x=0 and y=0
Ok,
Using the 2+x/1+y method we'd get x=2 and y=1(the original measures)
Using it with 2+x/1+y
2+2=4
1+1=2
But that doesn't passes through point (2,1)!
Here is the original solution, if it can help anyone.
https://lh4.googleusercontent.com/I...yyyyyyyyyy.JPG
Thanks.
Re: Finding the minimum area of triangle, coordinate plane
Hello, Zellator!
Your diagram is wrong.
Quote:
A line is drawn through (2,1), forming a right triangle in the 1st quadrant.
Determine the centroid of the triangle with minimum area. Code:


* 
B*
 * (2,1)
 o
 ♥ *
   +        *  
 A *

The directions could have been stated more clearly.
. . A line through (2,1) and the two axes form a right triangle in the 1st quadrant.
The line through (2,1) and slope $\displaystyle m$ has the equation:
. . $\displaystyle y  1 \:=\:m(x2) \quad\Rightarrow\quad y \:=\:mx + (22m)$
$\displaystyle \text{The }x\text{intercept is: }A\left(\tfrac{2m1}{m},\:0\right) \quad \hdots\text{ (the base of the right triangle)}$
$\displaystyle \text{The }y\text{intercept is: }B(0,\:12m) \quad\hdots\text{ (the height of the right triangle)}$
$\displaystyle \text{The area is: }\:A \;=\;\tfrac{1}{2}bh \;=\;\tfrac{1}{2}\left(\tfrac{2m1}{m}\right)(12m)$
. . . . . . . . . $\displaystyle A \;=\;2  2m  \tfrac{1}{2}m^{1}$
$\displaystyle \text{Minimize }A \quad\Rightarrow\quad m \,=\,\tfrac{1}{2}$
$\displaystyle \text{The vertices of the triangle are: }\:(0,\,0),\;(4,\,0),\;(0,\,2)$
Now locate the centroid of the triangle . . .
Re: Finding the minimum area of triangle, coordinate plane
Quote:
Originally Posted by
Soroban Hello, Zellator!
Your diagram is wrong.
The directions could have been stated more clearly.
. . A line through (2,1) and the two axes form a right triangle in the 1st quadrant.
The line through (2,1) and slope $\displaystyle m$ has the equation:
. . $\displaystyle y  1 \:=\:m(x2) \quad\Rightarrow\quad y \:=\:mx + (22m)$
$\displaystyle \text{The }x\text{intercept is: }A\left(\tfrac{2m1}{m},\:0\right) \quad \hdots\text{ (the base of the right triangle)}$
$\displaystyle \text{The }y\text{intercept is: }B(0,\:12m) \quad\hdots\text{ (the height of the right triangle)}$
$\displaystyle \text{The area is: }\:A \;=\;\tfrac{1}{2}bh \;=\;\tfrac{1}{2}\left(\tfrac{2m1}{m}\right)(12m)$
. . . . . . . . . $\displaystyle A \;=\;2  2m  \tfrac{1}{2}m^{1}$
$\displaystyle \text{Minimize }A \quad\Rightarrow\quad m \,=\,\tfrac{1}{2}$
Now locate the centroid of the triangle . . .
Hi Soroban!
Thanks for your reply!
That's why this looked so easy at first!
Of course!!!!
That's great to know!!!
Hahhaha yeah, it is easy now!
Thanks for your help Soroban and Also sprach Zarathustra!!
Centroid is in the point $\displaystyle \left(\frac{0+4+0}{3},\frac{0+0+2}{3}\right) \rightarrow \left(\frac{4}{3},\frac{2}{3}\right)$
I'll keep my head up if something like this comes again!
All the best!
Re: Finding the minimum area of triangle, coordinate plane
Quote:
Originally Posted by
Zellator Hi Also sprach Zarathustra!
Thanks for your reply!
That's great to know, as well.
An alternative way to do it is with 2+x and 1+y, as I previously mentioned.
As it turns out, m=1/2
So if we plug this into your equation we get x=0 and y=0
Ok,
Using the 2+x/1+y method we'd get x=2 and y=1(the original measures)
Using it with 2+x/1+y
2+2=4
1+1=2
But that doesn't passes through point (2,1)!
Here is the original solution, if it can help anyone.
https://lh4.googleusercontent.com/I...yyyyyyyyyy.JPG
Thanks.
It's the same solution!
You took (x,y) and (2,1) and used slope definition to get: {1+y}/{2+x} = m
...and so on and so on...
Re: Finding the minimum area of triangle, coordinate plane
Quote:
Originally Posted by
Also sprach Zarathustra It's the same solution!
You took (x,y) and (2,1) and used slope definition to get: {1+y}/{2+x} = m
...and so on and so on...
Yes, it is the same.
But if you don't know that this 4 and 2 are on the y and x axes it is difficult to consider them.
If you understand what I mean. :)
Thanks!