Question

Write the equation of the mirror line for the segment with endpoints (5, 4) and (-1, -1).

Answer 1

The equation of the mirror line in standard form is: 10y+12x=39. The reflection of the point is A'(1,2)

Step-by-step explanation:

Given that the endpoints of the segments are: (5,4) and (-1,-1) then the equation of the mirror line will be;

slope=m=Δy/Δx

Δy= -1 - 4= -5

Δx= -1-5 = -6

m= -5/-6 = 5/6

For the equation, taking point (5,4), and (x,y), it will be;

m=Δy/Δx

5/6= y-4/x-5

5(x-5) = 6(y-4)

5x-25 =6y-24

5x=6y-24+25

5x=6y+1

5x-1=6y

6y=5x-1 -------divide both sides by 6

y=5/6x -1/6 -------Equation of line segment

The mirror line is a perpendicular bisector of the segment.This means it passes at the midpoint of the segment.

Finding the midpoint of the segment will be;

{(x₁+x₂)/2 ,(y₁+y₂)/2}

{(5+-1)/2, (4+-1)/2} =(4/2, 3/2) =(2, 3/2)

Using the coordinate point of the midpoint, you can find the equation of the mirror line knowing that the slope will be -6/5 because the mirror line is perpendicular to the line segment that has a slope of 5/6.

Finding the equation of the mirror line using point (2, 3/2), (x,y) and m= -6/5

$\frac{y-1.5}{x-2} =\frac{-6}{5} \\\\5(y-1.5)=-6(x-2)\\5y-7.5=-6x+12\\5y-\frac{15}{2} =-6x+12\\\\10y-15=-12x+24\\\\10y+12x=15+24\\\\10y+12x=39$

The equation of the mirror line in standard form is;

10y+12x=39

2. Given point A(3,-2)  and line y=1/2x -1 ,you can use a graph tool to plot the equation for the mirror line, plot the object point and locate the image point across the mirror line.

From the graph, the image is A'(1,2)

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Reflection of a point :brainly.com/question/12865568

Keywords : equation, mirror line, segment, endpoints, reflection

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Answer 2

Answer:

  a) 12x +10y = 39

  b) A'(1, 2)

Step-by-step explanation:

a) The mirror line is the perpendicular bisector of the segment between the given points. The difference in their coordinates is ...

  (Δx, Δy) = (5, 4) -(-1, -1) = (6, 5)

The midpoint between the given points is ...

  ((5, 4) +(-1, -1))/2 = (2, 3/2)

One way to write the equation of the perpendicular line through point (h, k) is ...

  Δx(x -h) +Δy(y -k) = 0

Filling in the numbers from above, this becomes ...

  6(x -2) +5(y -3/2) = 0

  6x +5y -39/2 = 0 . . . . . . eliminate parentheses

  12x +10y = 39 . . . . . . . . . mirror line in standard form

__

b) The line through A perpendicular to the mirror line will have a slope that is the negative reciprocal of that of the mirror line: -1/(1/2) = -2. Then the point-slope equation of the line through A is ...

  y +2 = -2(x -3)

Using the equation of the mirror line to substitute for y, we can find the point on the mirror line that is the midpoint between A and its reflection.

  (1/2x -1) +2 = -2(x -3)

  1/2x +1 = -2x +6 . . . . . . eliminate parentheses

  5/2x = 5 . . . . . . . . . . . . add 2x-1

  x = 2 . . . . . . . . . . . . . . . multiply by 2/5

  y = (1/2)(2) -1 = 0 . . . . . find the y-coordinate of the midpoint using the equation for y

Now we know that (x, y) = (2, 0) is the midpoint between A and A'.

  (2, 0) = (A +A')/2 . . . . equation for midpoint of A and A'

  (4, 0) = A +A' . . . . . . . multiply by 2

  (4, 0) - A = A' = (4, 0) -(3, -2) . . . . subtract A

  A' = (1, 2) . . . . . the reflection of point A