Question
Using the distance formula, show taht the given points are collinear:
(-1, -1), (2, 3) and (8, 11)
(-1, -1), (2, 3) and (8, 11)
✓
Answer
Let A(-1, -1), B(2, 3) and C(8, 11) be the given points
Then,
$\text{AB}=\sqrt{(2-1)^2+(3+1)^2}=\sqrt{(3)^2+(4)^2}$
$=\sqrt{9+16}=\sqrt{25}=5\text{ units}$
$\text{BC}=\sqrt{(8-2)^2+(11-3)^2}=\sqrt{(6)^2+(8)^2}$
$=\sqrt{36+64}=\sqrt{100}=10\text{ units}$
$\text{AC}=\sqrt{(8+1)^2+(11+1)^2}=\sqrt{(9)^2+(12)^2}$
$=\sqrt{81+144}=\sqrt{225}=15\text{ units}$
$\therefore\text{AB}+\text{AC}=5+10=15\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
Then,
$\text{AB}=\sqrt{(2-1)^2+(3+1)^2}=\sqrt{(3)^2+(4)^2}$
$=\sqrt{9+16}=\sqrt{25}=5\text{ units}$
$\text{BC}=\sqrt{(8-2)^2+(11-3)^2}=\sqrt{(6)^2+(8)^2}$
$=\sqrt{36+64}=\sqrt{100}=10\text{ units}$
$\text{AC}=\sqrt{(8+1)^2+(11+1)^2}=\sqrt{(9)^2+(12)^2}$
$=\sqrt{81+144}=\sqrt{225}=15\text{ units}$
$\therefore\text{AB}+\text{AC}=5+10=15\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
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