Question

wednesday
draw △hij after a 90° rotation about the origin.
graph with h(-3, 3), i(1, 2), j(-2, 1)
thursday
give the vertices of a triangle that is 3 times as big as △abc if a(0, 0), b(2, 6), and c(6, 4). (centered at the origin)

Explanation:

Step1: 90° rotation rule (origin)

For a point $(x,y)$, 90° counterclockwise rotation about origin gives $(-y,x)$; clockwise gives $(y,-x)$. We use counterclockwise (standard):

• $H(-3,3) \to (-3,-3)$
• $I(1,2) \to (-2,1)$
• $J(-2,1) \to (-1,-2)$

Step2: Scale triangle by factor 3

Multiply each vertex of $\triangle ABC$ by scale factor 3:

• $A(0,0) \to (0 \times 3, 0 \times 3) = (0,0)$
• $B(2,6) \to (2 \times 3, 6 \times 3) = (6,18)$
• $C(6,4) \to (6 \times 3, 4 \times 3) = (18,12)$

Answer:

Wednesday (Rotated $\triangle HIJ$ vertices):

$H'(-3,-3)$, $I'(-2,1)$, $J'(-1,-2)$ (connect these points to draw the triangle)

Thursday (Scaled triangle vertices):

$A'(0,0)$, $B'(6,18)$, $C'(18,12)$