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mart [117]
2 years ago
15

Find equation of set of points pieces that its distance from the point 3, 4, -5 and -2, 1, 4 are equal.

Mathematics
1 answer:
Gnesinka [82]2 years ago
3 0

Answer:

Step-by-step explanation:

Suppose we a point P(x,y,z) such that its distance from either the point A(3,4,-5) or B(-2,1,4) is the same.

Using this information we can formula:

distance AP = distance BP

first, let's find the distance from AP, using the distance formula.

r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}

AP = \sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2}

similarly, we can find the distance BP

BP = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}

since both distances are exactly the same we can equate them

AP = BP

\sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2} = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}

we can simplify it a bit squaring both sides, and getting rid of the subscripts.

(3 - x)^2 + (4 - y)^2 + (-5 - z)^2 = (-2 - x)^2 + (1 - y)^2 + (4 - z)^2

what we have done here is formulated an equation which consists of any point P that will have the same distance from (3,4,-5) and (-2,1,4).

To put it more concretely,

This is the equation of the the plane from that consists of all points (P) from which the distance from both (3,4,-5) and (-2,1,4) are equal.

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A recipe calls for 2 1/3 cups of flour to make a batch of cookies. How many batches of cookies can be made if you have 11 2/3 cu
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Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
2 years ago
Find the value of x<br><br> 2+4( 3 + 2x ) = 3x + 8
solniwko [45]
2 + 4(3+2x) = 3x + 8

2 + 12 + 8x = 3x + 8

14 - 8 = 3x - 8x

6 = -5x

x = -6/5
8 0
2 years ago
Read 2 more answers
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