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ziro4ka [17]
3 years ago
14

A land surveyor places two stakes 500 ft apart and creates a perpendicular to the line that connects these two stakes. He needs

to place a third stake 100 ft away that is equidistant to the two original stakes. To apply the Perpendicular Bisector Theorem, the land surveyor would need to identify​
PLEASE HELP

Mathematics
1 answer:
aksik [14]3 years ago
8 0

Answer:

A perpendicular bisector is a perpendicular line that divides a line into two equal halves

step by step:

To apply the Perpendicular Bisector Theorem, the land surveyor will need to identify; the midpoint along the line connecting the two stakes

The reason why the surveyor needs to identify the midpoint is as follows:

According to the Perpendicular Bisector Theorem, all points on the perpendicular bisector of a line are of equal distance from the line's endpoint

In order to apply the perpendicular bisector, the surveyor would need to identify the midpoint from which the third stake can be placed 100 ft. along a perpendicular bisector drawn from the midpoint

Therefore, the surveyor needs to identify; the midpoint along the line connecting the two stakes

uhhhh

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(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca
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Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation det(A-\lambdaI)=0 where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6

So the characteristic polynomial is \lambda^2-5\lambda+6=0.

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0

So \lambda=3, \lambda=2

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If \lambda=3 we get the following matrix

\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right].

Since both rows are equal, we have the equation

x-2y=0. Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case \lambda=2, using the same process, we get the vector (1,1).

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P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0
3 years ago
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