Question

the graph shows quadrilaterals bcde and bcde. which of the following transformations maps bcde onto bcde? translation left 5 units and down 7 units translation left 6 units and down 5 units reflection across the x - axis reflection across the y - axis rotation 90° clockwise around the origin rotation 90° counterclockwise around the origin rotation 180° around the origin

Explanation:

Step1: Check translation

Take a point, say \(B(1,2)\). For a translation left \(5\) units and down \(7\) units, the new - coordinates would be \((1 - 5,2 - 7)=(-4,-5)\). But \(B'\) is not at \((-4,-5)\). For a translation left \(6\) units and down \(5\) units, the new - coordinates of \(B(1,2)\) would be \((1 - 6,2 - 5)=(-5,-3)\) which is the coordinate of \(B'\). Check other points: For \(C(4,2)\), after translation left \(6\) units and down \(5\) units, we get \((4 - 6,2 - 5)=(-2,-3)\) which is \(C'\). For \(D(2,8)\), we get \((2 - 6,8 - 5)=(-4,3)\) which is \(D'\). For \(E(1,6)\), we get \((1 - 6,6 - 5)=(-5,1)\) which is \(E'\).

Step2: Check reflection and rotation

For reflection across the \(x\) - axis, the \(y\) - coordinate changes sign. For example, \(B(1,2)\) would become \((1,-2)
eq B'\). For reflection across the \(y\) - axis, the \(x\) - coordinate changes sign. For example, \(B(1,2)\) would become \((-1,2)
eq B'\). For a \(90^{\circ}\) clock - wise rotation around the origin, the transformation rule is \((x,y)\to(y, - x)\). For \(B(1,2)\), it would be \((2,-1)
eq B'\). For a \(90^{\circ}\) counter - clockwise rotation around the origin, the transformation rule is \((x,y)\to(-y,x)\). For \(B(1,2)\), it would be \((-2,1)
eq B'\). For a \(180^{\circ}\) rotation around the origin, the transformation rule is \((x,y)\to(-x,-y)\). For \(B(1,2)\), it would be \((-1,-2)
eq B'\).

Answer:

translation left 6 units and down 5 units