Someone im stuck polygon abcd is dilated by a scale factor of 2 with the center of dilation at the origin to create polygon a′b′c′d′. if the endpoints of are located at (0, -7) and (8, 8), what is the length of ? use the distance formula to you decide:

The distance formula is given by  \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Take point A(0, -7) as (x₁, x₂) and point B(8, 8) as (y₁, y₂)

Distance AB =  \sqrt{(8-0)^2+(8--7)^2}

Distance AB =  \sqrt{8^2+15^2}

Distance AB =  \sqrt{289}

Distance AB = 17

The line AB is dilated to A'B' by scale factor of 2, it means the length of the line is multiplied by 2

Distance A'B' = 2 × 17 = 34 units

Hope this helps!

~LENA~

The distance formula is given by  \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Take point A(0, -7) as (x₁, x₂) and point B(8, 8) as (y₁, y₂)

Distance AB =  \sqrt{(8-0)^2+(8--7)^2}

Distance AB =  \sqrt{8^2+15^2}

Distance AB =  \sqrt{289}

Distance AB = 17

The line AB is dilated to A'B' by scale factor of 2, it means the length of the line is multiplied by 2

Distance A'B' = 2 × 17 = 34 units

So, the answer to your question is 34 units

i got 10.28 but that rounds up to 10.3