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2 years ago

The given line segment has a midpoint at (3,1)

Mathematics

2 answers:

masya89 [10]2 years ago

5 0

Answer:

$y=\frac{1}{3}x$

Step-by-step explanation:

We have been given an image of a line segment and we are asked to find the equation of perpendicular bisector of our given line segment in slope-intercept form.

Since we know that the slope of perpendicular line to a given line is negative reciprocal of the slope of the given line.

Let us find the slope of our given line using slope formula.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$ , where,

$y_2-y_1$ = Difference between two y-coordinates,

$x_2-x_1$ = Difference between two x-coordinates of same y-coordinates.

Upon substituting the coordinates of points (2,4) and (4,-2) in slope formula we will get,

$\text{Slope}=\frac{-2-4}{4-2}$

$\text{Slope}=\frac{-6}{2}$

$\text{Slope}=-3$

Now we will find negative reciprocal of $-3$ to get the slope of perpendicular line.

$\text{Negative reciprocal of }-3=-(-\frac{1}{3})=\frac{1}{3}$

Since point (3,1) lies on the perpendicular line, so we will substitute coordinates of point (3,1) in slope-intercept form of equation $(y=mx+b)$ .

$1=\frac{1}{3}\times 3+b$

$1=1+b$

$1-1=1-1+b$

$b=0$

Therefore, the equation of perpendicular line will be $y=\frac{1}{3}x+0=\frac{1}{3}x$ and option A is the correct choice.

KiRa [710]2 years ago

4 0

Answer:

Step-by-step explanation:

Edge 2020

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Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

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Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: