Given the vertices of δabc are a (2, -5,) b (-4, 6) and c (3, 1), find the vertices following each of the transformations from the original vertices :
(i honestly have no idea what to do and i really need with this one)

I. A' =  (5,-5), B' = (-1,6), C' = (6,1)

II. A' =  (5,-11), B' = (-1,0), C' = (6,-5)

III. A' = (-5,-2), B = (6,4), C' = (1,-3)

Step-by-step explanation:

We are given the vertices of ΔABC as A = (2,-5), B = (-4,6) and C = (3,1).

I. It is required to 'reflect the triangle about the line x= 3'.

This rule changes (x,y) to (x+3,y).

So, the new vertices are given by,

A' = (2+3,-5) = (5,-5)

B' = (-4+3,6) = (-1,6)

C' = (3+3,1) = (6,1)

II. It is required to 'translate the triangle 3 units to the right and 6 units down'.

This rule changes (x,y) to (x+3,y-6).

So, the new vertices are given by,

A' = (2+3,-5-6) = (5,-11)

B' = (-4+3,6-6) = (-1,0)

C' = (3+3,1-6) = (6,-5)

III. It is required to 'rotate the triangle by 90° about the origin counter-clockwise'.

This rule changes (x,y) to (y,-x).

So, the new vertices are given by,

A = (2,-5) implies A' = (-5,-2)

B = (-4,6) implies B = (6,4)

C = (3,1) implies C' = (1,-3)

c.35

step-by-step explanation:

8+13=21

21+8=29

29+6=35