MHB Area of Triangle Shaded Region

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The discussion revolves around calculating the area of a triangle using the determinant area formula, specifically for problem 21. The user is struggling to identify the necessary points in (x, y) format from a provided image. The area can be derived from the cross product of vectors formed by the triangle's vertices, with specific points given as B = (0,0), A = (-10,25), and C = (18,5). Additionally, there is a mention of the importance of understanding the underlying concepts rather than relying solely on external help. The conversation emphasizes the need for self-sufficiency in solving mathematical problems.
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Hello everyone. I am having trouble finding the area of the shaded region using the determinant area formula. I know where to plug in the numbers into the formula. My problem here is finding the needed points in the form (x, y) from the given picture for question 21.

Screenshot_20210115-185039_Drive.jpg
 
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Please only post one question per thread.

For the second problem what graphing utilities do have access to? (Assuming they don't want you to find one on the internet.)

-Dan
 
problem 21

let point B = (0,0)
A = (-10,25)
C = (18,5)

area of the triangle is half the area of the parallelogram formed by adjacent sides BC and BA which may be found using the cross product of two vectors ...

area = $\dfrac{1}{2}(\vec{BC} \times \vec{BA}) =
\dfrac{1}{2}\begin{vmatrix}
\vec{i} &\vec{j} & \vec{k}\\
18 & 5 & 0\\
-10 & 25 & 0
\end{vmatrix}=
\dfrac{1}{2}\begin{vmatrix}
18 & 5\\
-10 &25
\end{vmatrix}$
 
Last edited by a moderator:
topsquark said:
Please only post one question per thread.

For the second problem what graphing utilities do have access to? (Assuming they don't want you to find one on the internet.)

-Dan

I don't need help with 22. I just need the set up for 21.
 
skeeter said:
problem 21

let point B = (0,0)
A = (-10,25)
C = (18,5)

area of the triangle is half the area of the parallelogram formed by adjacent sides BC and BA which may be found using the cross product of two vectors ...

area = $\dfrac{1}{2}(\vec{BC} \times \vec{BA}) =
\dfrac{1}{2}\begin{vmatrix}
\vec{i} &\vec{j} & \vec{k}\\
18 & 5 & 0\\
-10 & 25 & 0
\end{vmatrix}=
\dfrac{1}{2}\begin{vmatrix}
18 & 5\\
-10 &25
\end{vmatrix}$

Can you tell me how you came up with the elements of the determinant?
 
Beer soaked ramblings follow.
mathland said:
Hello everyone. I am having trouble finding the area of the shaded region using the determinant area formula. I know where to plug in the numbers into the formula. My problem here is finding the needed points in the form (x, y) from the given picture for question 21.

View attachment 10940
Duplicate post from MHF by gufeliz (aka https://mathhelpboards.com/members/harpazo.8631/) that was deleted and who was subsequently banned by topsquark.

You really ought to read your book's relevant section before diving into the exercises.
And you should do it when you're fesh and full of energy; preferably after you've rested and slept (and presumably had some nourishment with coffee shortly afterwards) so that you can maximize your mental energy into understanding and applying what you've been reading and not when "my brain is tired and I am physically exhausted" as you like to embellish it. Regardless of how passionate you are about math, you can't work/study and concentrate as hard at the end of a study session (in your case, the end of a working day) as at the beginning.

mathland said:
I don't need help with 22. I just need the set up for 21.
Fourteen years of Precalculus review should have given you some insight on how to set up your problems on your own. How can you ever hope to fish on your own if you keep asking others to throw the net for you so you can just pull it up without knowing what made others throw the net the way they did.

Alternatively,

Vertices: (0, 25), (10, 0), (28, 5)

Area = $
\dfrac{1}{2}\begin{vmatrix}
\ 0& 25&1\\
10 & 0 & 1\\
28 & 5 & 1
\end{vmatrix}$
https://www.physicsforums.com/attachments/311846._xfImport
The area should be clear to you from this diagram.
If it's not then you really have a big problem.
 
Last edited:
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