Question

triangle abc was translated according to the rule ((x, y) \to (x + 1.5, y - 3.5)) to create the image (\triangle abc) shown on the coordinate plane. which graph shows the pre - image, (\triangle abc)?

Explanation:

Step1: Identify image coordinates

From the graph:
$A' = (0, 2)$, $B' = (0, 0)$, $C' = (-4, 0)$

Step2: Reverse the translation rule

The translation rule is $(x, y) \to (x+1.5, y-3.5)$. To find the pre-image, solve for original $(x,y)$:
$x = x' - 1.5$, $y = y' + 3.5$

Step3: Calculate pre-image points

For $A$: $x = 0 - 1.5 = -1.5$, $y = 2 + 3.5 = 5.5$ → $A(-1.5, 5.5)$
For $B$: $x = 0 - 1.5 = -1.5$, $y = 0 + 3.5 = 3.5$ → $B(-1.5, 3.5)$
For $C$: $x = -4 - 1.5 = -5.5$, $y = 0 + 3.5 = 3.5$ → $C(-5.5, 3.5)$

Answer:

The pre-image $\triangle ABC$ has vertices at $A(-1.5, 5.5)$, $B(-1.5, 3.5)$, and $C(-5.5, 3.5)$. When plotted, this is a right triangle positioned 1.5 units left and 3.5 units up from the image $\triangle A'B'C'$.