Find the distance between the points:

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)

$A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(15-9)^{2}+(11-3)^{2}}$

$=\sqrt{(15-9)^{2}+(11-3)^{2}}$

$=\sqrt{(6)^{2}+(8)^{2}}$

$=\sqrt{36+64}$

$=\sqrt{100}$

$=10$ 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)

$A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(-5-7)^{2}+\{1-(-4)\}^{2}}$

$=\sqrt{(-5-7)^{2}+(1+4)^{2}}$

$=\sqrt{(-12)^{2}+(5)^{2}}$

$=\sqrt{144+25}$

$=\sqrt{169}$

$=13$ 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)

$A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(9-(-6))^{2}+\{-12-(-4)\}^{2}}$

$=\sqrt{(9+6)^{2}+(-12+4)^{2}}$

$=\sqrt{(15)^{2}+(-8)^{2}}$

$=\sqrt{225+64}$

$=\sqrt{289}$

$=17$ 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)

$A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(4-1)^{2}+\{-6-(-3)\}^{2}}$

$=\sqrt{(4-1)^{2}+(-6+3)^{2}}$

$=\sqrt{(3)^{2}+(-3)^{2}}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=\sqrt{9 \times 2}$

$=3 \sqrt{2}$ 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)

$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{\{(a-b)-(a+b)\}^{2}+\{(a+b)-(a-b)\}^{2}}$

$=\sqrt{(a-b-a-b)^{2}+(a+b-a+b)^{2}}$

$=\sqrt{(-2 b)^{2}+(2 b)^{2}}$

$=\sqrt{4 b^{2}+4 b^{2}}$

$=\sqrt{8 b^{2}}$

$=\sqrt{4 \times 2 b^{2}}$

$=2 \sqrt{2} b$ units

$=\sqrt{\{(a-b)-(a+b)\}^{2}+\{(a+b)-(a-b)\}^{2}}$

$=\sqrt{(a-b-a-b)^{2}+(a+b-a+b)^{2}}$

$=\sqrt{(-2 b)^{2}+(2 b)^{2}}$

$=\sqrt{4 b^{2}+4 b^{2}}$

$=\sqrt{8 b^{2}}$

$=\sqrt{4 \times 2 b^{2}}$

$=2 \sqrt{2} b$ 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)

$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(a \cos \alpha-a \sin \alpha)^{2}+(-a \sin \alpha-a \cos \alpha)^{2}}$

$=\sqrt{\left(a^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha-2 a^{2} \cos \alpha \times \sin \alpha\right)+\left(a^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha+2 a^{2} \cos \alpha \times \sin \alpha\right)}$

$=\sqrt{2 a^{2} \cos ^{2} \alpha+2 a^{2} \sin ^{2} \alpha}$

$=\sqrt{2 a^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)}$

$=\sqrt{2 a^{2}(1)}$                      (From the identity $\cos ^{2} \alpha+\sin ^{2} \alpha=1$ )

$=\sqrt{2 a^{2}}$

$=\sqrt{2} a$ units