Find the distances between the following points.

(i) Given points are A(a,0) and B(0,a)
x₁ = a , y₁= 0, x₂ = 0, y₂ = a
By Distance formula,
d(A,B) = √[(x₂-x₁)²+(y₂-y₁)²]
d(A,B) = √[(0-a)²+(a-0)²]
= √(a²+a²)
= √(2a²)
= a√2 units.
Hence the distance between the points A and B is a√2 units.

(ii) Given points are P(-6,-3) and Q(-1,9)
x₁ = -6 , y₁= -3, x₂ = -1, y₂ = 9
By Distance formula,
d(P,Q) = √[(x₂-x₁)²+(y₂-y₁)²]
d(P,Q) = √[(-1-(-6))²+(9-(-3))²]
= √(5²+12²)
= √(25+144)
= √169
= 13
Hence the distance between the points P and Q is 13 units.

(iii) Given points are R(-3a, a) and S(a, -2a)
x₁ = -3a, y₁= a, x₂ = a, y₂ = -2a
By Distance formula,
d(R,S) = √[(x₂-x₁)²+(y₂-y₁)²]
d(R,S) = √[(a-(-3a))²+(-2a-a)²]
= √(4a)²+(-3a)²)
= √(16a²+9a²)
= √(25a²)
= 5a
Hence the distance between the points R and S is 5a units.