Question

triangle tvw is dilated according to the rule $d_{o, \frac{3}{4}}(x, y) \longrightarrow (\frac{3}{4}x, \frac{3}{4}y)$ to create the image triangle tvw, which is not shown. what are the coordinates of the endpoints of the segment tv? options: t(-1, 2) and v(0, 3); t(-3, 6) and v(0, 1); t(-1, 2) and v(0, 1); t(-3, 6) and v(0, 3)

Explanation:

Step1: Find coordinates of T and V

From the graph, T is at \((-4, 8)\) and V is at \((0, 4)\).

Step2: Apply dilation rule

The dilation rule is \(D_{O,\frac{3}{4}}(x, y)\to(\frac{3}{4}x, \frac{3}{4}y)\).
For T\((-4, 8)\):
\(x'\!=\!\frac{3}{4}(-4)\!=\! -3\)? Wait, no, wait—wait, maybe I misread T's coordinates. Wait, looking at the grid: T is at x = -4? Wait, no, the x - axis: from -6 to 4. Let's check again. Wait, the grid lines: T is at ( - 4, 8)? Wait, no, maybe T is at (-4, 8)? Wait, no, let's count the grid. Wait, the x - coordinate of T: from the origin (0,0), moving left 4 units? Wait, no, the x - axis labels: -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4. So T is at x = -4, y = 8? Wait, but when we apply the dilation \(\frac{3}{4}\), let's recalculate. Wait, maybe I made a mistake. Wait, the correct coordinates: Let's look at the graph again. T is at (-4, 8)? Wait, no, maybe T is at (-4, 8)? Wait, no, the user's graph: T is at ( - 4, 8)? Wait, no, let's check the options. Wait, the last option is T'(-3,6) and V'(0,3). Let's recalculate:
For T: if T is (-4, 8), then \(\frac{3}{4}(-4)= - 3\), \(\frac{3}{4}(8)=6\), so T'(-3,6).
For V: V is (0, 4), so \(\frac{3}{4}(0)=0\), \(\frac{3}{4}(4)=3\), so V'(0,3).
Ah, I see, I misread T's coordinates earlier. So T is (-4, 8), V is (0, 4).
Applying the dilation:
T': \(\frac{3}{4}(-4)= - 3\), \(\frac{3}{4}(8)=6\) → (-3, 6)
V': \(\frac{3}{4}(0)=0\), \(\frac{3}{4}(4)=3\) → (0, 3)

Answer:

T'(-3, 6) and V'(0, 3) (the last option: T'(-3, 6) and V'(0, 3))