BEGINNING COORDINATES: A- 2,2
B- 0,2
C- 4,4

Triangle ABC is dilated by a factor of 2 with the center of dilation at the origin to form triangle A′B′C′. What are the coordinates of triangle A′B′C′?
Question 1 options:

A)
A′ (0, 4), B′ (4, 0), C′ (16, 16)

B)
A′ (0, –4), B′ (–4, 0), C′ (–8, –8)

C)
A′ (0, 4), B′ (4, 0), C′ (8, 8)

D)
A′ (4, 0), B′ (0, 4), C′ (8, 8)

The correct option is;

A'(4, 4), B'(0, 4), C'(8, 8).

Step-by-step explanation:

The beginning coordinates are;

A -2, 2

B - 0, 2

C - 4, 4

Given that triangle ABC is dilated by 2, the center of dilation being the origin, (0, 0), we have;

Given that the center of dilation is the origin, we multiply the x, and y values by the scale factor as follows;

(x, y) → (2·x, 2·y)

Which gives;

A - (2, 2) → A'(2×2, 2×2) = A'(4, 4)

B - (0, 2) → B'(2×0, 2×2) = B'(0, 4)

C - (4, 4) → C'(2×4, 2×4) = C'(8, 8)

The correct option is A'(4, 4), B'(0, 4), C'(8, 8).

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

step-by-step explanation: