Question

test name: geometry module 04 - 2526 - 1206310
question 1 - 1 (interactions 1 & 2)
drop down item interactions
the transformation from △bag (pre - image) with coordinates b( - 2, - 9) a(9, - 7) g(3, - 5)
△bag (image) with coordinates b(14, - 9) a(3, - 7) g(9, - 5) is
described by a
reflection
rotation
of 90° clockwise about the origin
in the form (x,y)→(x + 6,y)
across the line y = - 6
across the line x = 6
translation
question 1 - 2 (interaction 3)
refer to the graph.

Explanation:

Step1: Check translation formula

For a translation \((x,y)\to(x + h,y + k)\), we check the \(x\)-coordinates and \(y\)-coordinates of the pre - image and image points.
For point \(B(-2,9)\) and \(B'(14,-9)\):

• For the \(x\)-coordinate: \(x\) of \(B\) is \(-2\), \(x\) of \(B'\) is \(14\). Using the formula \(x'=x + h\), then \(14=-2+h\), so \(h = 16\).
• For the \(y\)-coordinate: \(y\) of \(B\) is \(9\), \(y\) of \(B'\) is \(-9\). Using the formula \(y'=y + k\), then \(-9=9 + k\), so \(k=-18\). This is not the case.

For a rotation of \(90^{\circ}\) clockwise about the origin \((x,y)\to(y,-x)\). For \(B(-2,9)\), it would map to \((9,2)
eq(14,-9)\).

Step2: Check reflection formula

For a reflection across the line \(y = c\), the formula is \((x,y)\to(x,2c - y)\).
If \(c=-6\), for \(B(-2,9)\): \(x\) remains \(-2\), \(y\) becomes \(2\times(-6)-9=-12 - 9=-21
eq-9\).
For a reflection across the line \(x = c\), the formula is \((x,y)\to(2c - x,y)\).
If \(c = 6\), for \(B(-2,9)\): \(x\) becomes \(2\times6-(-2)=12 + 2=14\), \(y\) remains \(9\). But \(B'\) has \(y\)-coordinate \(-9\). Wait, no, we made a mistake. Wait, actually, for reflection across \(x = 6\):
For \(A(9,-7)\): \(x\) becomes \(2\times6-9=12 - 9 = 3\), \(y=-7\) (matches \(A'(3,-7)\)).
For \(B(-2,9)\): \(x\) becomes \(2\times6-(-2)=14\), \(y = 9\). But wait, no, the \(y\)-coordinates of \(B\) and \(B'\) are \(9\) and \(-9\). Wait, no, actually, we have two transformations: first reflection across \(x = 6\) (for \(x\)-coordinate transformation) and then reflection across \(y=-6\) (for \(y\)-coordinate transformation). But wait, no. Wait, let's check the \(x\)-coordinates:
For \(B(-2,9)\) and \(B'(14,-9)\), \(x\) transformation: \(x\to2\times6 - x\). For \(A(9,-7)\) and \(A'(3,-7)\), \(x\to2\times6 - x\). For \(G(3,-5)\) and \(G'(9,-5)\), \(x\to2\times6 - x\).
The \(y\)-coordinates: For \(B(9)\) and \(B'(-9)\), \(y\to2\times(-6)-y\). For \(A(-7)\) and \(A'(-7)\), \(y\to2\times(-6)-(-7)=-12 + 7=-5\) (no). Wait, no, actually, we observe that the \(x\)-coordinate transformation is \(x\to12 - x\) (reflection across \(x = 6\)) and \(y\)-coordinate remains the same for \(A\) and \(G\) in terms of \(y\)-value. Wait, no, \(A(9,-7)\to A'(3,-7)\): \(x\) changes from \(9\) to \(3\) (\(12-9 = 3\)), \(y=-7\) remains. \(B(-2,9)\to B'(14,-9)\): \(x = 12-(-2)=14\), \(y\) is not related by reflection across \(x = 6\) for \(y\). Wait, no, actually, we made a wrong start.
Let's check the \(x\)-coordinates:
The mid - point of \(B(-2)\) and \(B'(14)\) is \(\frac{-2 + 14}{2}=6\). The mid - point of \(A(9)\) and \(A'(3)\) is \(\frac{9+3}{2}=6\). The mid - point of \(G(3)\) and \(G'(9)\) is \(\frac{3 + 9}{2}=6\). So the transformation for \(x\)-coordinate is reflection across \(x = 6\) (since the line \(x = 6\) is the perpendicular bisector of the line segments joining pre - image and image points for \(x\)-coordinates).

Answer:

reflection, across the line \(x = 6\)