Using the distance formula, show that the given points are collinear:

Question:

Using the distance formula, show that the given points are collinear:
(i) (1, −1), (5, 2) and (9, 5)

(ii) (6, 9), (0, 1) and (−6, −7)
(iii) (−1, −1), (2, 3) and (8, 11) 

(iv) (−2, 5), (0, 1) and (2, −3)

Solution:

(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then

$A B=\sqrt{(5-1)^{2}+(2+1)^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5$ units

$B C=\sqrt{(9-5)^{2}+(5-2)^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5$ units

$A C=\sqrt{(9-1)^{2}+(5+1)^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{100}=10$ units

 

$\therefore A B+B C=(5+5)$ units $=10$ units $=A C$

Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then

$A B=\sqrt{(0-6)^{2}+(1-9)^{2}}=\sqrt{(-6)^{2}+(-8)^{2}}=\sqrt{100}=10$ units

$B C=\sqrt{(-6-0)^{2}+(-7-1)^{2}}=\sqrt{(-6)^{2}+(-8)^{2}}=\sqrt{100}=10$ units

$A C=\sqrt{(-6-6)^{2}+(-7-9)^{2}}=\sqrt{(-12)^{2}+(-16)^{2}}=\sqrt{400}=20$ units

$\therefore A B+B C=(10+10)$ units $=20$ units $=A C$

Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then

$A B=\sqrt{(2+1)^{2}+(3+1)^{2}}=\sqrt{(3)^{2}+(4)^{2}}=\sqrt{25}=5$ units

$B C=\sqrt{(8-2)^{2}+(11-3)^{2}}=\sqrt{(6)^{2}+(8)^{2}}=\sqrt{100}=10$ units

$A C=\sqrt{(8+1)^{2}+(11+1)^{2}}=\sqrt{(9)^{2}+(12)^{2}}=\sqrt{225}=15$ units

$\therefore A B+B C=(5+10)$ units $=15$ units $=A C$

Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then

$A B=\sqrt{(0+2)^{2}+(1-5)^{2}}=\sqrt{(2)^{2}+(-4)^{2}}=\sqrt{20}=2 \sqrt{5}$ units

$B C=\sqrt{(2-0)^{2}+(-3-1)^{2}}=\sqrt{(2)^{2}+(-4)^{2}}=\sqrt{20}=2 \sqrt{5}$ units

$A C=\sqrt{(2+2)^{2}+(-3-5)^{2}}=\sqrt{(4)^{2}+(-8)^{2}}=\sqrt{80}=4 \sqrt{5}$ units

$\therefore A B+B C=(2 \sqrt{5}+2 \sqrt{5})$ units $=4 \sqrt{5}$ units $=A C$

Hence, the given points are collinear.

 

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