Question

figure n is the result of a transformation on figure m. which transformation would accomplish this?
answer
a translation 1 unit left
a rotation 90° counterclockwise about the origin
a rotation 90° clockwise about the origin
a translation 1 unit right

Explanation:

Step1: Analyze translation

Translation moves a figure without rotation. Figure \(N\) is not just shifted left/right (as rotation is involved), so translation options can be ruled out.

Step2: Analyze rotation

For a \(90^{\circ}\) clockwise rotation about the origin \((x,y)\to(y, - x)\). For a \(90^{\circ}\) counter - clockwise rotation about the origin \((x,y)\to(-y,x)\). Observing the orientation of Figure \(M\) and Figure \(N\), a \(90^{\circ}\) clockwise rotation about the origin would not map Figure \(M\) to Figure \(N\).
If we consider a translation 1 unit left: assume a point \((x,y)\) on Figure \(M\), after translation 1 unit left it becomes \((x - 1,y)\). But the orientation (not just position shift) of Figure \(N\) relative to Figure \(M\) is a rotation.
If we consider a translation 1 unit right: assume a point \((x,y)\) on Figure \(M\), after translation 1 unit right it becomes \((x + 1,y)\). But the orientation (not just position shift) of Figure \(N\) relative to Figure \(M\) is a rotation.
Let's use a point - based approach. Suppose a vertex of Figure \(M\) is \((-1,0)\). After a \(90^{\circ}\) counter - clockwise rotation about the origin \((x,y)\to(-y,x)\), \((-1,0)\to(0, - 1)\) (not relevant). After a \(90^{\circ}\) clockwise rotation about the origin \((x,y)\to(y,-x)\), \((-1,0)\to(0,1)\) (not relevant).
Let's use the property of rigid transformations and visual inspection. If we consider the position of the figures on the coordinate - plane, a translation 1 unit left:
Take a reference point (say the right - most point of Figure \(M\) near \((-1,0)\)). If we move it 1 unit left, we get \((-2,0)\). But looking at the orientation (the "slant" of the figure), we can see that it is a rotation.
If we consider rotation:
The general form of a rotation. If we assume the center of rotation (for simplicity, assume rotation about a point that can be visually compared). By comparing the orientation of the two figures (the direction of the "slant" of the triangles), we can see that it is a \(90^{\circ}\) counter - clockwise rotation about the origin.

Answer:

A rotation \(90^{\circ}\) counterclockwise about the origin.