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nikitadnepr [17]
3 years ago
15

MATH HELP PLEASE

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

calculate the slopes of the lines using the gradient formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

(x₁, y₁ ) = A(- 4, - 1) and (x₂, y₂ ) = B(-1, 2 )

m_{AB} = \frac{2+1}{-1+4} = \frac{3}{3} = 1

(x₁, y₁ ) = B(-1, 2) and (x₂, y₂ ) = (5, 1)

m_{BC} = \frac{x1-2}{5+1} = - \frac{1}{6}

(x₁, y₁) = C(5, 1 ) and (x₂, y₂ ) = D(1, - 3)

m_{CD} = \frac{-3-1}{1-5} = \frac{-4}{-4} = 1

(x₁, y₁) = A(- 4, - 1) and (x₂, y₂) = D(1, - 3 )

m_{AD} = \frac{-3+1}{1+4} = - \frac{2}{5}

Quadrilateral ABCD is not a parallelogram since only one pair of opposite sides is parallel , that is AB and CD



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evablogger [386]

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Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

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Step # 1 : Swap the first and second matrix rows,

\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}

Step # 2 : Cancel leading coefficient in row 2 through R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1,

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Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

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As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0
3 years ago
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