Question
Verify, whether points $P(6,-6), Q(3,-7)$ and $R(3,3)$ are collinear.
✓
Answer
$\mathrm{PQ}=\sqrt{(6-3)^2+(-6+7)^2}$ by distance formula
$
\begin{aligned}
& =\sqrt{(3)^2+(1)^2}=\sqrt{10} \\
\mathrm{QR} & =\sqrt{(3-3)^2+(-7-3)^2} \\
& =\sqrt{(0)^2+(-10)^2}=\sqrt{100} \\
\mathrm{PR} & =\sqrt{(3-6)^2+(3+6)^2} \\
& =\sqrt{(-3)^2+(9)^2}=\sqrt{90} .
\end{aligned}
$
From I, II and III out of $\sqrt{10}, \sqrt{100}$ and $\sqrt{90}, \sqrt{100}$ is the largest number.
Now we will verify whether $(\sqrt{100})$ and $(\sqrt{10}+\sqrt{90})$ are equal.
For this compare $(\sqrt{100})^2$ and $(\sqrt{10}+\sqrt{90})^2$.
You will see that $(\sqrt{10}+\sqrt{90})>(\sqrt{100}) \therefore \mathrm{PQ}+\mathrm{PR} \neq \mathrm{QR}$
$\therefore$ points $\mathrm{P}(6,-6), \mathrm{Q}(3,-7)$ and $\mathrm{R}(3,3)$ are not collinear.
$
\begin{aligned}
& =\sqrt{(3)^2+(1)^2}=\sqrt{10} \\
\mathrm{QR} & =\sqrt{(3-3)^2+(-7-3)^2} \\
& =\sqrt{(0)^2+(-10)^2}=\sqrt{100} \\
\mathrm{PR} & =\sqrt{(3-6)^2+(3+6)^2} \\
& =\sqrt{(-3)^2+(9)^2}=\sqrt{90} .
\end{aligned}
$
From I, II and III out of $\sqrt{10}, \sqrt{100}$ and $\sqrt{90}, \sqrt{100}$ is the largest number.
Now we will verify whether $(\sqrt{100})$ and $(\sqrt{10}+\sqrt{90})$ are equal.
For this compare $(\sqrt{100})^2$ and $(\sqrt{10}+\sqrt{90})^2$.
You will see that $(\sqrt{10}+\sqrt{90})>(\sqrt{100}) \therefore \mathrm{PQ}+\mathrm{PR} \neq \mathrm{QR}$
$\therefore$ points $\mathrm{P}(6,-6), \mathrm{Q}(3,-7)$ and $\mathrm{R}(3,3)$ are not collinear.
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