**(i) **A (4, 8), B (5, 12), C (9, 28)

The slope of line AB =\(\dfrac{ 12 – 8 }{ 5 – 4}\)

= \( \dfrac{4}{1}\)

The slope of line BC = \(\dfrac{ 28 – 12 }{ 9 – 5}\)

= \( \dfrac{16}{4}\)

= 4

The slope of line CA =\(\dfrac{ 8 – 28}{4 – 9}\)

= \(\dfrac{ -20 }{ -5}\)

= 4

Here, AB = BC = CA

The Given points are collinear.

**(ii) **A(16, – 18), B(3, – 6), C(– 10, 6)

The slope of line AB = \( \dfrac{ 6 – (-18) }{ 3 – 16}\)

=\( \dfrac{12 }{ -13}\)

The slope of line BC =\( \dfrac{ 6 – (-6) }{ -10 – 3}\)

=\( \dfrac{12 }{ -13}\)

The slope of line CA = \(\dfrac{ 6 – (-18) }{ -10 – 16}\)

= \( \dfrac{12 }{ -13}\)

= 4

Here, AB = BC = CA

The Given points are collinear.

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