Two similar triangles are shown on the coordinate grid: Which set of transformations has been performed on triangle ABC to form triangle A′B′C′?

dilation by a scale factor of 4 followed by reflection about the x-axis
dilation by a scale factor of 2 followed by reflection about the x-axis
dilation by a scale factor of 4 followed by reflection about the y-axis
dilation by a scale factor of 2 followed by reflection about the y-axis

The set of transformations has been performed on triangle ABC is dilation by a scale factor of 2 followed by reflection about the x-axis

Step-by-step explanation:

Let us write the vertices of Δ ABC

∵ Vertex A = (-2, -1)

∵ Vertex B = (0, 0)

∵ Vertex C = (1, -3)

Let us Write the vertices of Δ A'B'C'

∵ Vertex A' = (-4, 2)

∵ Vertex B' = (0, 0)

∵ Vertex C' = (2, 6)

Let us find the scale factor k between the x-coordinates of the points and their images.

∵ -4 = k (-2) ⇒ Divide both side by -2

∵ =

∴ 2 = k

∴ The scale factor of a dilation is 2

∵ The signs of the y-coordinates of the images are the opposite of

    the signs of the y-coordinates of the points

∴ The triangle is reflected about the x-axis

∴ The set of transformations has been performed on triangle ABC

   is dilation by a scale factor of 2 followed by reflection about the x-axis

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

answer: 1 + 1 = 11

step-by-step explanation: