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