What is the slope of the line? A) 1 B) 6 C) 8 D) 1/2

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

8 0

8 because rise/run is 80/10

Digiron [165]3 years ago

7 0

C because when you count up the rise is 8 and the run is 1 so they cancel out and it becomes 8

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

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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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