Question

you can perform translations in the coordinate plane. translate $\triangle abc$ 4 units up to form $\triangle abc$. translate $\triangle abc$ 4 units up.

Explanation:

Step1: Identify coordinates of A, B, C

From the graph, let's assume the coordinates: Let's find the coordinates of each vertex. Let's say \( A(1, -2) \), \( B(4, -1) \), \( C(3, -3) \) (we can determine these by looking at the grid: x - coordinate (horizontal) and y - coordinate (vertical), with upward as positive y - direction and right as positive x - direction).

Step2: Apply translation rule (up 4 units)

The rule for translating a point \((x,y)\) \( k \) units up is \((x,y)\to(x,y + k)\). Here, \( k = 4 \).

• For point \( A(1,-2) \): New y - coordinate is \(-2+4 = 2\), so \( A'(1,2) \)
• For point \( B(4,-1) \): New y - coordinate is \(-1 + 4=3\), so \( B'(4,3) \)
• For point \( C(3,-3) \): New y - coordinate is \(-3 + 4 = 1\), so \( C'(3,1) \)

Step3: Plot the new points

Plot \( A'(1,2) \), \( B'(4,3) \), \( C'(3,1) \) on the coordinate plane and connect them to form \( \triangle A'B'C' \).

Answer:

To translate \( \triangle ABC \) 4 units up, we first identify the coordinates of \( A \), \( B \), and \( C \) (e.g., \( A(1, - 2) \), \( B(4, - 1) \), \( C(3, - 3) \)). Then we apply the translation rule \((x,y)\to(x,y + 4)\) to get \( A'(1,2) \), \( B'(4,3) \), and \( C'(3,1) \). Finally, we plot these new points and connect them to form \( \triangle A'B'C' \).