Write the equation of the mirror line for the segment with endpoints (5, 4) and (-1, -1).
find the reflection of point a(3, -2) through the mirror line of y = 1/2x − 1.

  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

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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

none cuz there are no "following statements" regarding the question

step-by-step explanation:

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.