9. Find the coordinates of each image point: a) \( D_{0,-2}:(1,3) \rightarrow(-2,-6) \) b) \( R_{o, 90^{\circ}}:(2,-4) \rightarrow(4,2) \) c) \( D_{(1,2), \frac{1}{2}}:(5,4) \rightarrow \) d) \( R_{(-2,-3),-90^{\circ}}:(2,-5) \rightarrow \)
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To find the coordinates of each image point after applying the transformations, we can break down each transformation step by step. For problem a, where we have a dilation \( D_{0,-2} \) (which scales by a factor of -2), we apply this to the point \((1,3)\): - Each coordinate is multiplied by \(-2\): - \(x\): \(1 \times -2 = -2\) - \(y\): \(3 \times -2 = -6\) So, the image point is \((-2, -6)\), which matches the given transformation. For problem b, the point is rotated \(R_{0, 90^{\circ}}\) around the origin \((2, -4)\): - The rotation rule for \(90^{\circ}\) counterclockwise around the origin changes the coordinates \((x, y)\) to \((-y, x)\): - For \( (2, -4) \): - New \(x\): \(-(-4) = 4\) - New \(y\): \(2\) Thus, the image point is \((4, 2)\). For problem c, we apply a dilation \(D_{(1,2), \frac{1}{2}}\) (from the point \((1,2)\) scaling by factor \(\frac{1}{2}\)): - The new coordinates will be: - \(x' = 1 + \frac{1}{2}(5 - 1) = 1 + 2 = 3\) - \(y' = 2 + \frac{1}{2}(4 - 2) = 2 + 1 = 3\) So, the image point is \((3, 3)\). For problem d, we rotate the point \((2, -5)\) by \(-90^{\circ}\) around \((-2, -3)\): - To perform this transformation, we first translate the point to the origin by subtracting \((-2, -3)\): - Translate \((2, -5)\) to \((2 - (-2), -5 - (-3)) = (4, -2)\). - Now we apply the \(-90^{\circ}\) rotation (which transforms \((x,y)\) to \((y,-x)\)): - For \((4, -2)\), the rotated coordinates are \((-2, -4)\). - Then we translate back: - Final coordinates: \((-2 - 2, -4 - 3) = (-4, -7)\). So, the answers to the transformations are: c) \((3, 3)\) d) \((-4, -7)\)