Question
Prove that the points (-2, 5), (0, 1) and (2, -3) are collinear.
✓
Answer
Let the points be A(-2, 5), B(0, 1) and C(2, -3)
Now $\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(0+2)^2+(1-5)^2}$
$=\sqrt{(2)^2+(-4)^2}=\sqrt{4+16}$
$=\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$
Similarly, $\text{BC}=\sqrt{(2-0)^2+(-3-1)^2}$
$=\sqrt{(2)^2+(-4)^2}$
$=\sqrt{4+16}=\sqrt{20}$
$=\sqrt{4\times5}=2\sqrt{5}$
$\text{CA}=\sqrt{(-2-2)^2+(5+3)^2}$
$=\sqrt{(-4)^2+(8)^2}=\sqrt{16+64}$
$=\sqrt{80}=\sqrt{16\times5}=4\sqrt{5}$
Now, $\text{AB}+\text{BC}=2\sqrt{5}+2\sqrt{5}$
and $\text{CA}=4\sqrt{5}$
AB + BC = CA
A, B and C are collinear.
Now $\text{AB}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(0+2)^2+(1-5)^2}$
$=\sqrt{(2)^2+(-4)^2}=\sqrt{4+16}$
$=\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$
Similarly, $\text{BC}=\sqrt{(2-0)^2+(-3-1)^2}$
$=\sqrt{(2)^2+(-4)^2}$
$=\sqrt{4+16}=\sqrt{20}$
$=\sqrt{4\times5}=2\sqrt{5}$
$\text{CA}=\sqrt{(-2-2)^2+(5+3)^2}$
$=\sqrt{(-4)^2+(8)^2}=\sqrt{16+64}$
$=\sqrt{80}=\sqrt{16\times5}=4\sqrt{5}$
Now, $\text{AB}+\text{BC}=2\sqrt{5}+2\sqrt{5}$
and $\text{CA}=4\sqrt{5}$
AB + BC = CA
A, B and C are collinear.
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