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Svetllana [295]
3 years ago
13

write an equation of the perpendicular bisector of the line segment whose endpoints are (-1,1) and (7,-5)

Mathematics
1 answer:
gladu [14]3 years ago
3 0
Find midpoint of segment.
(-1,1), (7,-5)
x= \frac{-1+7}{2}=3,y= \frac{1+(-5)}{2}=-2

Find slope of line which passes through (-1,1) and (7,-5).
m= \frac{-5-1}{7-(-1)}= \frac{-6}{8}=- \frac{3}{4}

The slope of perpendicular line is negative reciprocal value of m.
m_{\perp}=- \frac{1}{m} =- \frac{1}{- \frac{3}{4} } = \frac{4}{3}

Write quation of line which slope is \frac{4}{3} and passes throug the point (3,-2).
m=\frac{4}{3},x_1=3,y_1=-2
\\y-y_1=m(x-x_1)
\\
\\y-(-2)=\frac{4}{3}(x-3)
\\
\\y+2=\frac{4}{3}x-\frac{8}{3}
\\
\\3y+6=4x-8
\\4x-3y-14=0

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An aircraft on a reconnaissance mission takes off from its home base and flies 550 miles at a bearing of S 46° E to a location i
Alexandra [31]
<h3>Answer: D) 1692 miles</h3>

============================================

Explanation:

A diagram is very handy for this type of problem. Refer to the diagram below (attached image). Here are the steps used to form the diagram

  1. Plot point A and make that the home base. Plot point D directly south of point A. I'm skipping B and C because those will be used to form the triangle later.
  2. Imagine you are standing at point A and looking south at point D. Turn 46 degrees toward the east direction (this is what the notation "bearing of S 46° E" means). Then move 550 miles from A to land on point B. This is the location where the plane makes its first turn.
  3. Plot the points E and F, which are north and south of point B.
  4. Imagine we are at point B and looking directly south at point F. Turn 55 degrees toward the west, which is due to the "S 55° W" and then move 483 miles to arrive at point C. This is the plane's location where it turns around to go back to home base.

The diagram below shows all of this visually summarized. We have the segments

  • AB = 550 miles
  • BC = 483 miles
  • AC = x miles

And we have these angles

  • Angle DAB = 46 degrees and Angle ABE = 46 degrees (blue)
  • Angle ABC = 79 degrees (red)
  • Angle ACB = 55 degrees and Angle FBC = 55 degrees (green)

Angle ABC comes from the fact that the blue and green angles add with the red angle to get 180. So we basically need to solve 46+y+55 = 180 which leads to y = 79.

------------------------------

Once the diagram is set up, we'll use the law of cosines to find x.

Focus on triangle ABC. Ignore points D,E,F. Ignore any extra lines as well.

The interior angles of triangle ABC are

  • A = 46 degrees
  • B = 79 degrees
  • C = 55 degrees

The sides opposite the angles are

  • a = 483 (opposite angle A)
  • b = x (opposite angle B)
  • c = 550 (opposite angle C)

Now we can apply the law of cosines

b^2 = a^2 + c^2 - 2*a*c*cos(B)

x^2 = 483^2 + 550^2 - 2*483*550*cos(79)

x^2 = 434412.180756441

x = sqrt(434412.180756441)

x = 659.099522649229

x = 659

Side AC, or CA, is roughly 659 miles long.

------------

To find the total distance the plane travels, we add up the three sides of the triangle (ie we find the perimeter)

Distance Traveled = AB + BC + CA

Distance Traveled = 550 + 483 + 659

Distance Traveled = 1692 miles

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