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) \( R_{(-2,-3),-90^{\circ}}:(2,-5) \rightarrow \)
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
Mind Expander
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) \((-4, -7)\)