Question

26)
a) translation: 1 unit right and 2 units up
b) rotation 180° about the origin
c) rotation 90° clockwise about the origin
d) reflection across the x - axis

Explanation:

Step1: Analyze Option A (Translation)

Check if moving each vertex of triangle \( UVT \) 1 unit right and 2 units up matches \( U'V'T' \). Let's take point \( U \): suppose \( U \) has coordinates (let's assume grid coordinates, say \( U \) is at \( (-1, -1) \)? Wait, better to look at the graph. Wait, original triangle \( UVT \): \( U \) is above x-axis? Wait no, original \( U \) is below? Wait the graph: \( U \) is on the lower triangle, \( U' \) on upper. Wait, let's take coordinates. Let's assume each grid is 1 unit. Let's find coordinates of \( U \), \( V \), \( T \) and \( U' \), \( V' \), \( T' \).

Suppose \( U \) is at \( (-1, -1) \)? Wait no, looking at the graph: lower triangle \( UVT \): \( U \) is at (let's say) \( x=-1 \), \( y=-1 \)? Wait no, the x-axis is horizontal, y-axis vertical. Let's see: \( U \) is on the lower triangle, \( U' \) is on the upper. Let's check reflection over x-axis: reflection over x-axis changes \( (x,y) \) to \( (x, -y) \). Wait, no: reflection over x-axis is \( (x,y) \to (x, -y) \). Wait, but if \( U \) is at \( (-1, -1) \), reflection over x-axis would be \( (-1, 1) \). Wait, but \( U' \) is above x-axis. Wait, maybe better to check each option.

Option D: reflection across x-axis. Let's take point \( T \): suppose \( T \) is at \( (-1, -3) \), reflection over x-axis would be \( (-1, 3) \), which matches \( T' \)? Wait, no, the upper triangle \( T' \) is above. Wait, maybe I got the axes wrong. Wait, the y-axis is upward, x-axis rightward. So lower triangle: \( T \) is at the bottom, \( U \) above \( T \), \( V \) to the left. Upper triangle \( T' \) is above \( U' \), \( V' \) to the left. So reflection across x-axis: a point \( (x,y) \) becomes \( (x, -y) \). So if \( T \) is at \( (x, y) \) below x-axis, \( T' \) would be at \( (x, -y) \) above. Let's check coordinates. Let's assume \( U \) is at \( (-1, 1) \)? No, maybe I messed up. Wait, the key is: reflection across x-axis flips the y-coordinate. Let's check the triangles: the lower triangle (UVT) and upper (U'V'T') are mirror images across the x-axis? Wait no, the x-axis is between them. Wait, no: the x-axis is the horizontal line. The lower triangle is below? No, the x-axis is the middle line. Wait, the lower triangle is below the x-axis? No, the x-axis is the horizontal line, so above x-axis is positive y, below negative. Wait, the lower triangle \( UVT \): \( U \) is at (let's say) \( x=-1 \), \( y=-1 \), \( T \) at \( x=-1 \), \( y=-3 \), \( V \) at \( x=-2 \), \( y=-2 \). Then reflection over x-axis would be \( U'(-1, 1) \), \( T'(-1, 3) \), \( V'(-2, 2) \), which matches the upper triangle \( U'V'T' \). So that's reflection across x-axis. Let's check other options.

Option A: translation 1 right, 2 up. If \( U \) is at \( (-1, -1) \), translation 1 right, 2 up: \( (0, 1) \), but \( U' \) is at \( (-1, 1) \) (assuming), so no. Option B: rotation 180 about origin: \( (x,y) \to (-x, -y) \). So \( U(-1, -1) \) would become \( (1, 1) \), which doesn't match \( U'(-1, 1) \). Option C: rotation 90 clockwise about origin: \( (x,y) \to (y, -x) \). So \( U(-1, -1) \) would become \( (-1, 1) \)? Wait, no: 90 clockwise is \( (x,y) \to (y, -x) \). So \( (-1, -1) \to (-1, 1) \). Wait, that's same as reflection? No, wait, maybe my coordinates are wrong. Wait, maybe the correct transformation is reflection across x-axis. Let's confirm: reflection across x-axis: each point (x,y) becomes (x, -y). So if original \( U \) is (x, y) below x-axis, \( U' \) is (x, -y) above. Which matches the graph. So the correct option is D.

Answer:

D. reflection across the x - axis