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Tpy6a [65]
3 years ago
13

Use cross products to find the area of the triangle in the xy-plane defined by (1, 2), (3, 4), and (−7, 7).

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
8 0

I love these. It's often called the Shoelace Formula. It actually works for the area of any 2D polygon.


We can derive it by first imagining our triangle in the first quadrant, one vertex at the origin, one at (a,b), one at (c,d), with (0,0),(a,b),(c,d) in counterclockwise order.


Our triangle is inscribed in the a \times d rectangle. There are three right triangles in that rectangle that aren't part of our triangle. When we subtract the area of the right triangles from the area of the rectangle we're left with the area S of our triangle.


S = ad - \frac 1 2 ab -  \frac 1 2 cd - \frac 1 2 (a-c)(d-b) = \frac 1 2(2 ad - ab -cd - ad +ab +cd -bc) = \frac 1 2(ad -bc)


That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) will have the same area as one whose vertex C is translated to the origin.


We set a=x_1 - x_3, b= y_1  - y_3, c=x_2 - x_3, d=y_2- y_3


S= ad-bc=(x_1 - x_3)(y_2 - y_3) -(x_2-x_3)(y_1 - y_3)


That's a perfectly useful formula right there. But it's usually multiplied out:


S= x_1y_2 - x_1 y_3  - x_3y_2 + x_3 y_3 - x_2 y_1 + x_2y_3 + x_3 y_1 - x_3 y_3


S= x_1 y_2 - x_2 y_1  + x_2y_3 - x_3y_2   + x_3 y_1 - x_1 y_3


That's the usual form, the sum of cross products. Let's line up our numbers to make it easier.


(1, 2), (3, 4), (−7, 7)

(−7, 7),(1, 2), (3, 4),


[tex]A = \frac 1 2 ( 1(7)-2(-7) + 3(2)-4(1) + -7(4) - (7)(3)

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Step-by-step explanation:

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How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

\displaystyle\sum_{n=1}^N (21n-20) = 2332

Expand the sum on the left as

\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

and recall the formulas,

\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2

So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that

21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332

Solve for <em>N</em> :

21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664

21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

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