Question
Using the distance formula, show taht the given points are collinear:
(1, -1), (5, 2) and (9, 5)
(1, -1), (5, 2) and (9, 5)
✓
Answer
Let A(1, -1), B(5, 2) and C(9, 5) be the given points
Then,
$\text{AB}=\sqrt{(1-5)^2+(-1-2)^2}=\sqrt{(4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{BC}=\sqrt{(5-9)^2+(2-5)^2}=\sqrt{(-4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{AC}=\sqrt{(1-9)^2+(-1-5)^2}=\sqrt{(-8)^2+(-6)^2}$
$=\sqrt{64+36}=\sqrt{100}=10\text{ units}$
$\therefore\text{AB}+\text{AC}=5+5=10\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
Then,
$\text{AB}=\sqrt{(1-5)^2+(-1-2)^2}=\sqrt{(4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{BC}=\sqrt{(5-9)^2+(2-5)^2}=\sqrt{(-4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{AC}=\sqrt{(1-9)^2+(-1-5)^2}=\sqrt{(-8)^2+(-6)^2}$
$=\sqrt{64+36}=\sqrt{100}=10\text{ units}$
$\therefore\text{AB}+\text{AC}=5+5=10\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
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