Question

1. graph and label the figure and its image under the sequence of transformations. triangle lmn with vertices ( l(-3,-1) ), ( m(-3,-4) ), and ( n(-5,-8) ): a) ( 90^{circ} ) clockwise rotation about the origin b) reflected across the line ( x = 1 )

Explanation:

Step1: 90° clockwise rotation formula

For a point \((x,y)\) rotated \(90^{\circ}\) clockwise about the origin, the new coordinates \((x',y')\) are given by \((x',y')=(y, -x)\).

• For \(L(-3,-1)\): \(x=-3,y = - 1\), then \(x'=-1,y' = 3\), so \(L'(-1,3)\)
• For \(M(-3,-4)\): \(x=-3,y=-4\), then \(x'=-4,y' = 3\), so \(M'(-4,3)\)
• For \(N(-5,-8)\): \(x=-5,y=-8\), then \(x'=-8,y' = 5\), so \(N'(-8,5)\)

Step2: Reflection across the line \(x = 1\) formula

The formula for reflecting a point \((x,y)\) across the line \(x=a\) is \((x',y')=(2a - x,y)\). Here \(a = 1\), so \((x',y')=(2\times1-x,y)=(2 - x,y)\)

• For \(L'(-1,3)\): \(x=-1,y = 3\), then \(x'=2-(-1)=3,y' = 3\), so \(L''(3,3)\)
• For \(M'(-4,3)\): \(x=-4,y = 3\), then \(x'=2-(-4)=6,y' = 3\), so \(M''(6,3)\)
• For \(N'(-8,5)\): \(x=-8,y = 5\), then \(x'=2-(-8)=10,y' = 5\), so \(N''(10,5)\)

Answer:

The vertices of the original triangle \(L(-3,-1)\), \(M(-3,-4)\), \(N(-5,-8)\). After \(90^{\circ}\) clock - wise rotation about the origin: \(L'(-1,3)\), \(M'(-4,3)\), \(N'(-8,5)\). After reflection across \(x = 1\): \(L''(3,3)\), \(M''(6,3)\), \(N''(10,5)\). Plot the points \(L(-3,-1)\), \(M(-3,-4)\), \(N(-5,-8)\) for the original triangle, \(L'(-1,3)\), \(M'(-4,3)\), \(N'(-8,5)\) for the rotated triangle and \(L''(3,3)\), \(M''(6,3)\), \(N''(10,5)\) for the final reflected triangle on the coordinate grid.