Question

triangle mnp is dilated according to the rule $d_{o,1.5}(x,y) \to (1.5x, 1.5y)$ to create the image triangle mnp, which is not shown. what are the coordinates of the endpoints of segment mn? \bigcirc m(-6, 9) and n(4, 9) \bigcirc m(-6, 9) and n(3, 9) \bigcirc m(-2, 3) and n(7, 9) \bigcirc m(-2, 3) and n(1, 3)

Explanation:

Step1: Find coordinates of M and N

From the graph, M is at (-4, 6) and N is at (2, 6).

Step2: Apply dilation rule

Dilation rule: \((x,y) \to (1.5x, 1.5y)\).
For M(-4, 6): \(x = -4 \times 1.5 = -6\), \(y = 6 \times 1.5 = 9\) → \(M'(-6, 9)\).
For N(2, 6): \(x = 2 \times 1.5 = 3\)? Wait, no, wait: Wait, 21.5 is 3? Wait, no, the options have N'(3,9) or N'(4,9)? Wait, wait, maybe I misread N's coordinate. Wait, looking at the graph, N is at (2,6)? Wait, no, maybe the grid: Let me check again. Wait, the x-axis: from -6 to 4. M is at (-4,6), N is at (2,6)? Wait, no, maybe N is at (2,6)? Wait, but when we do 1.52=3, but the first option has N'(4,9). Wait, maybe I made a mistake. Wait, maybe M is (-4,6) and N is (2,6)? Wait, no, let's recalculate. Wait, 1.5(-4) = -6, 1.56=9 (so M'(-6,9)). For N: if N is at (2,6), 1.52=3, 1.56=9 → N'(3,9). But wait, the first option is M'(-6,9) and N'(4,9). Wait, maybe N is at (8/3,6)? No, the graph: Let's look at the grid. The x-coordinate of N: from the graph, N is at (2,6)? Wait, no, maybe the original N is at (4/3,6)? No, the grid lines: each square is 1 unit. So M is at (-4,6), N is at (2,6). Then dilation by 1.5: (-41.5, 61.5)=(-6,9), (21.5,61.5)=(3,9). So the second option: M'(-6,9) and N'(3,9). Wait, but let's check the options again. The options are:

1. M'(-6,9) and N'(4,9)
2. M'(-6,9) and N'(3,9)
3. M'(-2,3) and N'(7,9)
4. M'(-2,3) and N'(1,3)

Wait, maybe I misread N's coordinate. Let's check the graph again. The triangle MNP: M is at (-4,6), N is at (2,6)? Wait, no, maybe N is at (8/3,6)? No, the grid: each x-tick is 1. So from -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4. So M is at (-4,6), N is at (2,6). Then 1.52=3, so N'(3,9). So the second option is M'(-6,9) and N'(3,9). But wait, let's check the first option: M'(-6,9) and N'(4,9). 1.5x=4 → x=4/1.5=8/3≈2.666, which is not an integer. So maybe N is at (8/3,6)? No, the graph shows N at (2,6). So the correct calculation: M(-4,6) → (-6,9), N(2,6) → (3,9). So the second option: M'(-6,9) and N'(3,9).

Answer:

M'(-6, 9) and N'(3, 9) (the second option: M'(-6, 9) and N'(3, 9))