Find the distance between the points:

(i) A(9, 3) and B(15, 11)
(ii) A(7, −4) and B(−5, 1)
(iii) A(−6, −4) and B(9, −12)
(iv) A(1, −3) and B(4, −6)
(v) P(a + b, a − b) and Q(a −b, a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)

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Solution

(i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x₁ = 9, y₁ = 3) and (x₂ = 15, y₂ = 11)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(15 - 9)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(15 - 9)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units$

(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x₁= 7, y₁ = −4) and (x₂= −5, y₂ = 1)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(- 5 - 7)}^{2} + {\{1 - (- 4)\}}^{2}} = \sqrt{{(- 5 - 7)}^{2} + {(1 + 4)}^{2}} = \sqrt{{(- 12)}^{2} + {(5)}^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13 units$

(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x₁ = −6, y₁ = −4) and (x₂ = 9, y₂ = −12)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(9 - (- 6))}^{2} + {\{- 12 - (- 4)\}}^{2}} = \sqrt{{(9 + 6)}^{2} + {(- 12 + 4)}^{2}} = \sqrt{{(15)}^{2} + {(- 8)}^{2}} = \sqrt{225 + 64} = \sqrt{289} = 17 units$

(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x₁= 1, y₁ = −3) and (x₂ = 4, y₂ = −6)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(4 - 1)}^{2} + {\{- 6 - (- 3)\}}^{2}} = \sqrt{{(4 - 1)}^{2} + {(- 6 + 3)}^{2}} = \sqrt{{(3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2} units$

(v) P(a + b, a − b) and Q(a −b, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x₁ = a + b, y₁ = a − b) and (x₂ = a − b, y₂ = a + b)
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{\{(a - b) - (a + b)\}}^{2} + {\{(a + b) - (a - b)\}}^{2}} = \sqrt{{(a - b - a - b)}^{2} + {(a + b - a + b)}^{2}} = \sqrt{{(- 2 b)}^{2} + {(2 b)}^{2}} = \sqrt{4 b^{2} + 4 b^{2}} = \sqrt{8 b^{2}} = \sqrt{4 \times 2 b^{2}} = 2 \sqrt{2} b units$

(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x₁ = a sin α, y₁ = a cos α) and (x₂ = a cos α, y₂ = −a sin α)
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(a \cos α - a \sin α)}^{2} + {(- a \sin α - a \cos α)}^{2}} = \sqrt{(a^{2} \cos^{2} α + a^{2} \sin^{2} α - 2 a^{2} \cos α \times \sin α) + (a^{2} \sin^{2} α + a^{2} \cos^{2} α + 2 a^{2} \cos α \times \sin α)} = \sqrt{2 a^{2} \cos^{2} α + 2 a^{2} \sin^{2} α} = \sqrt{2 a^{2} (\cos^{2} α + \sin^{2} α)} = \sqrt{2 a^{2} (1)} (From the identity \cos^{2} α + \sin^{2} α = 1) = \sqrt{2 a^{2}} = \sqrt{2} a units$

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Find the distance between the points: (i) A(9, 3) and B(15, 11) (ii) A(7, −4) and B(−5, 1) (iii) A(−6, −4) and B(9, −12) (iv) A(1, −3) and B(4, −6) (v) P(a + b, a − b) and Q(a −b, a + b) (vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)

Find the distance between the points:

(i) A(9, 3) and B(15, 11)
(ii) A(7, −4) and B(−5, 1)
(iii) A(−6, −4) and B(9, −12)
(iv) A(1, −3) and B(4, −6)
(v) P(a + b, a − b) and Q(a −b, a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)