What is the equation of a line that passes through the points (3, 6) and (8, 4)

1 answer:

4 0

First let us find the slope between the two points.

Slope = Change in y / Change in x.

(3, 6) and (8, 4) compares to (x₁, y₁) and (x₂, y₂)

Slope = (y₂ - y₁) / (x₂ - x₁) = (4 - 6) / (8 - 3) = -2/5

Slope m = -2/5= -0.4

Using y = mx + c, using point (3, 6)

y = -0.4x + c

6 = -0.4*3 + c

6 = -1.2 + c

6 + 1.2 = c

7.2 = c

c = 7.2

Equation = : y = -0.4x +c

y = -0.4x + 7.2

y = -4/10 + 72/10

10y = -4x + 72

5y = -2x + 36

Equation =: 5y = -2x + 36

Hope this explains it.

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Answer:

$||\overrightarrow{v}||\approx14.64$

Step-by-step explanation:

The magnitude of a vector $\displaystyle \overrightarrow{v}=\langle x,y\rangle$ is $||\overrightarrow{v}||=\sqrt{x^2+y^2}$ , which essentially describes the length of the vector itself between the initial and terminal points:

$||\overrightarrow{v}||=\sqrt{x^2+y^2}\\\\||\overrightarrow{v}||=\sqrt{(10.5)^2+(10.2)^2}\\\\||\overrightarrow{v}||=\sqrt{110.25+104.04}\\\\||\overrightarrow{v}||=\sqrt{214.29}\\\\||\overrightarrow{v}||\approx14.64$