Given that: 15 cot A = 8
⇒ cot A = 8/15
Let cot A = 8k/15k, where k is a real number.
In ∆ ABC, by Pythagoras theorem, we have
AC² = AB² + BC²
= (8k)²+(15k)²
= 64k²+225k²
= 289k²
⇒ AC = √(289k²) = 17k
Hence, sin A = BC/AC = 15k/17k = 15/17 and sec A = AC/AB = 17k/8k = 17/8

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(i) A(2,3), B(4,1)
Using the distance formula: √((x₂-x₁)²+(y₂-y₁)²)
Distance between A(2,3) and B(4,1) is given by AB
= √((4-2)²+(1-3)²) = √(4+4) = √8 = 2√2

(ii) P(-5,7), Q(-1,3)
Using the distance formula: √((x₂-x₁)²+(y₂-y₁)²)
Distance between P(-5,7) and Q(-1,3) is given by PQ
= √((-1-(-5))²+(3-7)²) = √(16+16) = √32 = 4√2

(iii) M(a,b), N(-a,-b)
Using the distance formula: √((x₂-x₁)²+(y₂-y₁)²)
Distance between M(a,b) and N(-a,-b) is given by MN
= √(-a-(a))²+(-b-(-b))²) = √(4a²+4b²) = 2√(a²+b²)

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Here, A(1,5), B(2, 3) and C(-2,-11).
Using distance formula: √((x₂-x₁)²+(y₂-y₁)²)
AB = √((2-1)²+(3-5)²) = √(1+4) = √5
BC = √((-2-2)²+(-11-3)²) = √(16+196) = √212
CA = √((1-(-2))²+(5-(-11))² = √(9+256) = √265
Here, AB+BC = √5+√212 ≠ √265 = AC
Hence, the points A(1,5), B(2,3) and C(-2,-11) are not collinear.