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GrogVix [38]
3 years ago
13

find an equation for the perpendicular bisector of the line segment whose endpoints are (-8,-5) and (-4,1)

Mathematics
2 answers:
saw5 [17]3 years ago
7 0

Answer:

y=\frac{3}{2}x-3

Step-by-step explanation:

<u>Step 1: Perpendicular bisector</u>

To find the perpendicular bisector of the segment, apply the midpoint formula:

\bigg(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\bigg)

Points: {(-8, -5, (-4, 1)}

x₁ = -8      first x value

x₂ = -4     second x value

y₁ = -5     first y value

y₂ = 1      second y value

Plug the points into the formula:

\bigg(\frac{-8+(-4)}{2}, \frac{-5+1}{2}\bigg)

Solve:

\bigg(\frac{-8+(-4)}{2}, \frac{-5+1}{2}\bigg)

=\bigg(\frac{-12}{2}, \frac{-4}{2}\bigg)

=(-6,-2)

The midpoint is (-6, -2).

<u>Step 2: Slope</u>

To find the slope (m), apply the formula:

\frac{y_2-y_1}{x_2-x_1}

(point location is the same as previous step)

Plug the points into the formula; then solve:

\frac{1-(-5)}{-4(-8)}

=\frac{6}{4}

m=\frac{3}{2}

The slope is 3/2

<u>Step 3: Solving for b</u>

y = mx+b

-6=\frac{3}{2}(-2)+b

\frac{3}{2}\left(-2\right)+b=-6

-\frac{3}{2}\cdot \:2+b=-6

-3+b=-6

-3+b+3=-6+3

b=-3

Therefore, the equation is \bold{-6=\frac{3}{2}(-2)-3}

lbvjy [14]3 years ago
6 0

Answer:

y = -x - 8

Step-by-step explanation:

Midpoint of the line:

x = (x1 + x2)/2 = (-8 + -4)/2 = -12/2 = -6

y = (y1 + y2)/2 = (-5 + 1)/2 = -4/2 = -2

so midpoint is (-6,-2)

Slope of the line: Slope m = (y2-y1)/(x2-x1)

m = (-1 - -5)/(-4 - -8) = (-1 + 5)/(-4 + 8) = 4/4 = 1

Perpendicular lines have slopes that are negative reciprocals of one another

so slope of the perpendicular line is -1/1 is -1

y = mx + b

y = -x + b

Using (-6,-2)

-2 = -(-6) + b

-2 = 6 + b

b = -8

so y = -x - 8

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In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the<br> function with its graph
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Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

f(x)=(x-2)^{2}+1

45)

The function g(x) is given by:

g(x)=f(x-1)

using f(x) we can find f(x-1)

g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

g(x)'=2(x-3)

2(x-3)=0

x=3

Puting it on g(x) we will get y value.

y=g(3)=(3-3)^{2}+1

y=g(3)=1

<u>Then, the minimum point of this function is (3,1) and it corresponds to (A)</u>

46)

Let's use the same method here.

g(x)=f(x+2)

g(x)=((x+2)-2)^{2}+1

g(x)=(x)^{2}+1

Let's find the first derivative and equal to zero to find x and y minimum value.

g'(x)=2x

0=2x

x=0

Evaluatinf g(x) at this value of x we have:

g(0)=(x)^{2}+1

g(0)=1

<u>Then, the minimum point of this function is (0,1) and it corresponds to (C)</u>

47)

Let's use the same method here.

g(x)=f(x)+2

g(x)=(x-2)^{2}+1+2

g(x)=(x-2)^{2}+3

Let's find the first derivative and equal to zero to find x and y minimum value.

g'(x)=2(x-2)

0=2(x-2)

x=2

Evaluatinf g(x) at this value of x we have:

g(2)=(2-2)^{2}+3

g(2)=3

<u>Then, the minimum point of this function is (2,3) and it corresponds to (B)</u>

48)

Let's use the same method here.

g(x)=f(x)-3

g(x)=(x-2)^{2}+1-3

g(x)=(x-2)^{2}-2

Let's find the first derivative and equal to zero to find x and y minimum value.

g'(x)=2(x-2)

0=2(x-2)

x=2

Evaluatinf g(x) at this value of x we have:

g(2)=(2-2)^{2}-2

g(2)=-2

<u>Then, the minimum point of this function is (2,-2) and it corresponds to (D)</u>

<u />

I hope it helps you!

<u />

8 0
3 years ago
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