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If the points $\mathrm{A}(1,2), \mathrm{B}(2,4)$ and $C(3, \mathrm{a})$ are collinear, what is the length $\mathrm{BC}$ ?
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The correct answer is:
$\sqrt{5}$ unit
Since the points are collinear.
$\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 4 & 1 \\ 3 & a & 1\end{array}\right|=0$
Expanding the determinant
$\Rightarrow 1\left|\begin{array}{ll}4 & 1 \\ \mathrm{a} & 1\end{array}\right|-2\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|+1\left|\begin{array}{ll}2 & 4 \\ 3 & \mathrm{a}\end{array}\right|=0$
$\Rightarrow(4-a)-2(2-3)+1(2 a-12)=0$
$\Rightarrow \quad 4-a+2+2 a-12=0$
$\Rightarrow \quad a-6=0$
$\Rightarrow \quad a=6$
Thus, Coordinates of $\mathrm{C}$ are $(3,6)$. Thus, $\mathrm{BC}=\sqrt{(3-2)^{2}+(6-4)^{2}}$
$=\sqrt{1+4}=\sqrt{5}$ unit
$\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 4 & 1 \\ 3 & a & 1\end{array}\right|=0$
Expanding the determinant
$\Rightarrow 1\left|\begin{array}{ll}4 & 1 \\ \mathrm{a} & 1\end{array}\right|-2\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|+1\left|\begin{array}{ll}2 & 4 \\ 3 & \mathrm{a}\end{array}\right|=0$
$\Rightarrow(4-a)-2(2-3)+1(2 a-12)=0$
$\Rightarrow \quad 4-a+2+2 a-12=0$
$\Rightarrow \quad a-6=0$
$\Rightarrow \quad a=6$
Thus, Coordinates of $\mathrm{C}$ are $(3,6)$. Thus, $\mathrm{BC}=\sqrt{(3-2)^{2}+(6-4)^{2}}$
$=\sqrt{1+4}=\sqrt{5}$ unit
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