Triangle A″B″C″ is formed by a reflection over y = −3 and dilation by a scale factor of 2 from the origin. Which equation shows the correct relationship between ΔABC and ΔA″B″C″? coordinate plane with triangle ABC at A negative 3 comma 3, B 1 comma negative 3, and C negative 3 comma negative 3

Option (3)

Step-by-step explanation:

This question is not complete; here is the complete question.

Triangle A″B″C″ is formed by a reflection over y = −3 and dilation by a scale factor of 2 from the origin. Which equation shows the correct relationship between ΔABC and ΔA″B″C″?

Coordinates of the vertices of the triangle ABC are,

A(-3, 3), B(1, -3) and C(-3, -3)

When triangle ABC is reflected over y = -3

Coordinates of the image triangle A'B'C' will be.

A(-3, 3) → A'(-3, -9)

B(1, -3) → B'(1, -3)

C(-3, -3) → C'(-3, -3)

Further ΔA'B'C' is dilated by a scale factor of 2 about the origin then the new vertices of image triangle A"B"C" will be,

Rule for the dilation will be,

(x, y) → (kx, ky) [where 'k' is the scale factor]

A'(-3, -9) → A"(-6, -18)

B'(1, -3) → B"(2, -6)

C'(-3, -3) → C"(-6, -6)

Length of AB =

                      =

                      =

                      =

Length of A"B" =

                         =

                         =

                         =

Therefore,

Option (3) is the answer.

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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tangent

step-by-step explanation: