Answer:
Explanation:
Coordinates of the vertices of ΔABC,
A(-6, -1), B(-3, -3), C(-1, -2)
Step - 1
Rule for the translation of a point (x, y) by 'h' units right and 'k' units upwards,
A(x, y) → A(x + h, y + k)
If ΔABC is translated by 5 units right and 2 units up, image points will be,
A(-6, -1) → H(-6 + 5, -1 + 2)
→ H(-1, 1)
B(-3, -3) → J(-3 + 5, -3 + 2)
→ J(2, -1)
C(-1, -2) → K(-1 + 5, -2 + 2)
→ K(4, 0)
Step - 2
If the image triangle HJK is reflected across a line
, rule for the reflection will be,
H(x, y) → A'(-y, -x)
By this rule,
H(-1, 1) → A'(-1, 1)
J(2, -1) → B'(1, -2)
K(4, 0) → C'(0, -4)
Step - 3
Rule for the rotation of a point 270° counterclockwise about the origin,
A'(x, y) → A"(y, -x)
By this rule, image points of ΔA'B'C' will be,
A'(-1, 1) → A"(1, 1)
B'(1, -2) → B"(-2, -1)
C'(0, -4) → C"(-4, 0)
Now we can graph the image triangle A"B"C".