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zmey [24]
3 years ago
12

Draw the triangle with vertices a(1, 1), b(5, 3), c(1, 7). find the parametrization, including endpoints, and sketch to check. (

enter your answers as a comma-separated list of equations. let x and y be in terms of t.)
Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0
Part A:

The parametric equations x=x_1+(x_2-x_1)t and y=y_1+(y_2-y_1)t where 0\leq t\leq1 describes the line segment that joins the points (x_1,\ y_1) and (x_2,\ y_2).

The parameterization of the line joining points A(1, 1) and B(5, 3) are given by substituting x_1=1,\ \ y_1=1,\ \ x_2=5,\ \ y_2=3.

Thus, we have

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

Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 + 4t and y = 1 + 2t.



Part B:

The parametric equations x=x_1+(x_2-x_1)t and y=y_1+(y_2-y_1)t where 0\leq t\leq1 describes the line segment that joins the points (x_1,\ y_1) and (x_2,\ y_2).

The parameterization of the line joining points B(5, 3) and C(1, 7) are given by substituting x_1=5,\ \ 
y_1=3,\ \ x_2=1,\ \ y_2=7.

Thus, we have

x = 5 + (1 - 5)t = 5 - 4t
y = 3 + (7 - 3)t = 3 + 4t

Therefore, the parameterization of line joining points B(5, 3) and C(1, 7) are x = 5 - 4t and y = 3 + 4t.



Part C:

The parametric equations x=x_1+(x_2-x_1)t and y=y_1+(y_2-y_1)t where 0\leq t\leq1 describes the line segment that joins the points (x_1,\ y_1) and (x_2,\ y_2).

The parameterization of the line joining points C(1, 7) and A(1, 1) are given by substituting x_1=1,\ \ 
y_1=7,\ \ x_2=1,\ \ y_2=1.

Thus, we have

x = 1 + (1 - 1)t = 1
y = 7 + (1 - 7)t = 7 - 6t

Therefore, the parameterization of line joining points A(1, 1) and B(5, 3) are x = 1 and y = 7 - 6t.
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Step-by-step explanation:

Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.

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Let A = 4, B = 5 and C = 6

A(B-C) = 4 \times (5-6) = 4 \times -1 = -4

Let A = 4, B = 6 and C = 5

A(B-C) = 4 \times (6-5) = 4 \times 1 = 4

Let A = 5, B =4 and C = 6

A(B-C) = 5 \times (4-6) = 5 \times -2 = -10

Let A = 5, B = 6 and C = 4

A(B-C) = 4 \times (6-4) = 4 \times 2 = 8

Let A = 6, B = 4 and C = 5

A(B-C) = 6 \times (4-5) = 6 \times -1 = -6

Let A = 6, B = 5 and C = 4

A(B-C) = 6 \times (6-5) = 6 \times 1 = 6

Summarizing the above values in the form of a table:

\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\  4 & 6 & 5 & 4\\  5 & 4 & 6 & -10\\  5 & 6 & 4 & 10\\  6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}

So, the least possible result is <em>-10</em>.

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