Question
Using the distance formula, show taht the given points are collinear:
(-2, 5), (0, 1) and (2, -3).
(-2, 5), (0, 1) and (2, -3).
✓
Answer
Let A(-2, 5), B(0, 1) and C(2, -3) be the given points
Then,
$\text{AB}=\sqrt{(0-2)^2+(1-5)^2}=\sqrt{(2)^2+(-4)^2}$
$=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\text{ units}$
$\text{BC}=\sqrt{(2+0)^2+(-3-1)^2}=\sqrt{(2)^2+(-4)^2}$
$=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\text{ units}$
$\text{AC}=\sqrt{(2+2)^2+(-3-5)^2}=\sqrt{(4)^2+(-8)^2}$
$=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}\text{ units}$
$\therefore\text{AB}+\text{AC}=2\sqrt{5}+2\sqrt{5}=4\sqrt{5}\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
Then,
$\text{AB}=\sqrt{(0-2)^2+(1-5)^2}=\sqrt{(2)^2+(-4)^2}$
$=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\text{ units}$
$\text{BC}=\sqrt{(2+0)^2+(-3-1)^2}=\sqrt{(2)^2+(-4)^2}$
$=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\text{ units}$
$\text{AC}=\sqrt{(2+2)^2+(-3-5)^2}=\sqrt{(4)^2+(-8)^2}$
$=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}\text{ units}$
$\therefore\text{AB}+\text{AC}=2\sqrt{5}+2\sqrt{5}=4\sqrt{5}\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
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