Question

give the new coordinates for dilating triangle lmn with vertices l(2, -8), m(12, 8), and n(14, -4): k = \frac{1}{2}
write the numerical answer (ie if you get 2 for your answer, type \2\ not \two\)
l ( type your answer... , type your answer... )
m ( type your answer... , type your answer... )
n ( type your answer... , type your answer... )

Explanation:

Step1: Recall dilation rule

To dilate a point \((x, y)\) with a scale factor \(k\), the new coordinates are \((k \cdot x, k \cdot y)\).

Step2: Dilate point L(2, -8)

For \(L(2, -8)\) and \(k = \frac{1}{2}\), we calculate \(x\)-coordinate: \(\frac{1}{2} \times 2 = 1\), \(y\)-coordinate: \(\frac{1}{2} \times (-8) = -4\). So \(L'(1, -4)\).

Step3: Dilate point M(12, 8)

For \(M(12, 8)\) and \(k = \frac{1}{2}\), \(x\)-coordinate: \(\frac{1}{2} \times 12 = 6\), \(y\)-coordinate: \(\frac{1}{2} \times 8 = 4\). So \(M'(6, 4)\).

Step4: Dilate point N(14, -4)

For \(N(14, -4)\) and \(k = \frac{1}{2}\), \(x\)-coordinate: \(\frac{1}{2} \times 14 = 7\), \(y\)-coordinate: \(\frac{1}{2} \times (-4) = -2\). So \(N'(7, -2)\).

Answer:

\(L'(1, -4)\)
\(M'(6, 4)\)
\(N'(7, -2)\)