Question

give the new coordinates for dilating square mnop with vertices m(-7, -4), n(-4, -3), o(-3, -6), and p(-6, -7): k = 2
there is no picture for this question.
write the numerical answer (ie if you get 2 for your answer, type \2\ not \two\)
m ( type your answer... , type your answer... )
n ( type your answer... , type your answer... )
o ( type your answer... , type your answer... )
p ( type your answer... , type your answer... )

Explanation:

Step1: Recall dilation rule

To dilate a point \((x, y)\) with a scale factor \(k\) centered at the origin, the new coordinates \((x', y')\) are given by \(x' = k \cdot x\) and \(y' = k \cdot y\).

Step2: Dilate point M(-7, -4)

For \(M(-7, -4)\) and \(k = 2\):
\(x' = 2\times(-7)= -14\)
\(y' = 2\times(-4)= -8\)
So \(M'(-14, -8)\)

Step3: Dilate point N(-4, -3)

For \(N(-4, -3)\) and \(k = 2\):
\(x' = 2\times(-4)= -8\)
\(y' = 2\times(-3)= -6\)
So \(N'(-8, -6)\)

Step4: Dilate point O(-3, -6)

For \(O(-3, -6)\) and \(k = 2\):
\(x' = 2\times(-3)= -6\)
\(y' = 2\times(-6)= -12\)
So \(O'(-6, -12)\)

Step5: Dilate point P(-6, -7)

For \(P(-6, -7)\) and \(k = 2\):
\(x' = 2\times(-6)= -12\)
\(y' = 2\times(-7)= -14\)
So \(P'(-12, -14)\)

Answer:

\(M'(-14, -8)\)
\(N'(-8, -6)\)
\(O'(-6, -12)\)
\(P'(-12, -14)\)