Question

rotate the figure 270° around (-5, -4). k(-13, -3) l(-7, 3) m(-6, -2)

Explanation:

Step1: Translate points relative to center

First, translate the points \(K(-13,-3)\), \(L(-7,3)\), \(M(-6,-2)\) so that the center of rotation \((- 5,-4)\) becomes the origin. Subtract the coordinates of the center of rotation from the coordinates of each point.
For point \(K\): \((-13-(-5),-3 - (-4))=(-8,1)\)
For point \(L\): \((-7-(-5),3-(-4))=(-2,7)\)
For point \(M\): \((-6 - (-5),-2-(-4))=(-1,2)\)

Step2: Apply 270 - degree rotation rule

The rule for a 270 - degree counter - clockwise rotation about the origin \((x,y)\to(y,-x)\).
For the translated point of \(K(-8,1)\), the rotated point is \((1,8)\)
For the translated point of \(L(-2,7)\), the rotated point is \((7,2)\)
For the translated point of \(M(-1,2)\), the rotated point is \((2,1)\)

Step3: Translate points back

Add the coordinates of the original center of rotation \((-5,-4)\) to the rotated points.
For the rotated and translated point of \(K\): \((1+(-5),8+(-4))=(-4,4)\)
For the rotated and translated point of \(L\): \((7+(-5),2+(-4))=(2,-2)\)
For the rotated and translated point of \(M\): \((2+(-5),1+(-4))=(-3,-3)\)

Answer:

\(K'(-4,4)\)
\(L'(2,-2)\)
\(M'(-3,-3)\)