Question

triangle abc will be dilated according to the rule ( d_{f, 0.25}(x,y) ), where point f is the center of dilation. what will be the coordinates of vertex a of the image? (-8,-4) (-2,-1) (0,0) (1,0)

Explanation:

Step1: Find coordinates of A and F

From the graph, \( A(-5, -3) \), \( F(3, 1) \).

Step2: Apply dilation formula

The dilation rule \( D_{F, 0.25}(x, y) \) means we first find the vector from \( F \) to \( A \), scale it by \( 0.25 \), then add back to \( F \).
Vector \( \overrightarrow{FA} = (-5 - 3, -3 - 1) = (-8, -4) \).
Scaled vector: \( 0.25 \times (-8, -4) = (-2, -1) \).
New coordinate \( A' = F + \) scaled vector: \( (3 - 2, 1 - 1) = (1, 0) \)? Wait, no—wait, correction: Wait, dilation from center \( F \): the formula is \( A' = F + 0.25(A - F) \).
So \( A'_x = 3 + 0.25(-5 - 3) = 3 + 0.25(-8) = 3 - 2 = 1 \).
\( A'_y = 1 + 0.25(-3 - 1) = 1 + 0.25(-4) = 1 - 1 = 0 \). Wait, but let's check again. Wait, maybe I misread A's coordinates. Wait, looking at the graph: A is at (-5, -3)? Wait, no, let's re - examine the grid. The x - axis: from left, A is at x = -5? Wait, the grid lines: each square is 1 unit. Let's see: F is at (3,1). A: let's count the squares. From F (3,1), moving left 8 units (x: 3 - 8 = -5) and down 4 units (y: 1 - 4 = -3). So A is (-5, -3). Then the vector from F to A is (-8, -4). Scaling by 0.25: (-2, -1). Then adding to F: (3 - 2, 1 - 1) = (1, 0). Wait, but let's check the options. Option D is (1,0). Wait, but maybe my initial coordinate of A is wrong. Wait, maybe A is at (-5, -3)? Wait, let's check the triangle. B is at (-5,1)? No, B is at (-5,1)? Wait, no, the y - axis: B is at y = 1? Wait, the red triangle: A is at (-5, -3), B is at (-5,1), C is at (1, -1). Then F is at (3,1). Then vector FA: A - F = (-5 - 3, -3 - 1)=(-8, -4). Multiply by 0.25: (-2, -1). Then A' = F + 0.25*(A - F) = (3 - 2, 1 - 1)=(1,0). So the coordinates of A' are (1,0).

Answer:

D. (1, 0)