I WILL GIVE BRAINLIEST Draw the line of reflection that reflects △ABC\triangle ABC△ABCtriangle, A, B, C onto △A′B′C′\triangle A'B'C'△A′B′C′triangle, A, prime, B, prime, C, prime.

I WILL GIVE BRAINLIEST Draw the line of reflection that reflects ABCtriangle ABCABCtriangle A B C onto ABCtriangle ABCABCtriangle A prime B prime C prime class=

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Answer:

See diagram attached.

Step-by-step explanation:

The line of reflection of preimage ABC reflected onto image A'B'C' can be found by the perpendicular bisector of lines joining the corresponding points of the image and preimage, for example, AA', or BB'.

See diagram attached.

Ver imagen mathmate

The equation of  Line of reflection is y + 2x =6

What is section formula?

When a point on a line segment divides it into two segments, the formula used to determine the coordinates of that point is known as the section formula.

Let us say, we have a point P(x,y) that divides the line segment with marked points as A (x1,y1) and B(x2,y2). To find the coordinates, we use the section formula, which is mathematically expressed as:

P= [tex](\frac{(mx_2 + nx_1} {2} . \frac{mx_1 + nx_2}{2} )[/tex]

It is given that,

  • A=(-3,2)
  • A'=(5,6),
  • B=(2,-3)
  • and B'=(6,-1),

As, the mid point of vertices of figure to the corresponding vertices of  reflected figure should lie on reflected line.

Since, Δ ABC is reflected to Δ A'B'C' through this line,

Let, Q be the mid point of  line AA', using section formula

Coordinate of Q = [tex](\frac{(-3) + 5}{2} . \frac{6+2}{2} ) = (1. 4 )[/tex]

and R be the mid point of line BB', using section formula

Coordinate of R = [tex](\frac{2 + 6}{2} . \frac{-3-1}{2} ) = (4. -2 )[/tex]

As Q and R lies on the line of the reflection,

Thus, the equation of the line of reflection,

[tex](y-4)=\frac{-2-4} {4-1} (x-1)[/tex]

3(y-4) = -6 (x-1)

3y - 12= 6 - 6x

3y + 6x = 18

y + 2x =6

The reflected line can be drawn by finding the solution of y+ 2x =6.

Learn more Section Formula here:

https://brainly.com/question/12103512

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