Respuesta :
Firstly, we will draw diagram of the scenario
Dana reflects point A(2,5) across line small script letter lℓ to get image point Upper A primeA′(6,1)
It means that AA' is perpendicular bisects of line(l)
so, let's assume it intersects at point M
so, M will be mid-point between A and A'
[tex]M=(\frac{2+6}{2},\frac{5+1}{2})[/tex]
[tex]M=(4,3)[/tex]
M is a point on line (l)
now, we will find slope of line on AA'
x1=2 , y1=5
x2=6 , y2=1
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
now, we can plug values
[tex]m=\frac{1-5}{6-2}[/tex]
[tex]m=-1[/tex]
now, we know that line (l) and AA' are perpendicular to each other
so, slope of line(l) is -1/m
slope=1
so, we have slope and a point
[tex]m=1[/tex]
[tex]M=(4,3)[/tex]
now, we can use point slope form of line
[tex]y-y_1=m(x-x_1)[/tex]
now, we can plug values
[tex]y-3=1(x-4)[/tex]
[tex]y=x-1[/tex]
so, the equation of line (l) is
[tex]y=x-7[/tex]..............Answer
Answer:
[tex]x-y=1[/tex]
Step-by-step explanation:
Given,
The point A(2, 5) is reflected by line and form the point A'(6, 1),
Since, the point of reflection is equidistant from the point and its image after reflection,
Hence, the coordinate of the point of reflection ( lies on the line of reflection ) are,
[tex](\frac{2+6}{2},\frac{5+1}{2})[/tex]
[tex]=(\frac{8}{2}, \frac{6}{2})[/tex]
= (4, 3)
Now, the slope of a line having end points A and A'
[tex]m=\frac{1-5}{6-2}[/tex]
[tex]=\frac{-4}{4}[/tex]
= - 1,
If slope of the line of reflection is m',
[tex]\implies m'\times m = -1[/tex] ( the line of reflection and the line joining the point and its image are perpendicular )
[tex]m'\times -1=-1\implies m'=1[/tex]
Thus, the line of reflection passes through the point (4, 3) and has the slope -1,
Hence, the equation of line of reflection,
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-3=1(x-4)[/tex]
[tex]y-3=x-4[/tex]
[tex]x-y=1[/tex]
