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3 years ago

Find the midpoint of the line segment whose endpoints are (2.6,5.1) and (3,4.7).

Mathematics

1 answer:

svp [43]3 years ago

7 0

Answer:

Mid point is: (2.8;4.9)

Step-by-step explanation:

To find midpoint of a line segment we can use the general equation:

$Mid=\frac{x_1+x_2}{2} ;\frac{y_1+y_2}{2}$

Where the point of the line are: (x₁;y₁) and (x₂;y₂).

In the problem, x₁ = 2.6, y₁ = 5.1 and x₂ = 3 and y₂ = 4.7. Replacing in the equation:

$Mid=\frac{2.6+3}{2} ;\frac{5.1+4.7}{2}$

<h3>Mid point is: (2.8;4.9)</h3>

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Express the system as AX = B; then solve using matrix inverses

Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

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$A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]$

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Adj.A=Transpose of cofactor of Matrix A

$Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2$

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]$

$\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4$

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