What is the standard form of this complex number?
2 answers:
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Answer:
Step-by-step explanation:
6g + 4a = 7a
⇔ 6g = 7a - 4a
⇔ 6g = 3a
Divide both side by 6a
⇒ ⇔
I hope this will help you.
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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