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The middle point of $\mathrm{A}(1,2)$ and $\mathrm{B}(\mathrm{x}, \mathrm{y})$ is $\mathrm{C}(2,4)$. If BD is perpendicular to $\mathrm{AB}$ such that $\mathrm{CD}=3$ umit, then what is the length BD?
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The correct answer is:
2 unit
Given that mid point of $\mathrm{A}(1,2)$ and $\mathrm{B}(\mathrm{x}, \mathrm{y})$ is $\mathrm{C}(2,4)$,
$\frac{1+x}{2}=2$ and $\frac{2+y}{2}=4$
$\Rightarrow \quad x=3$ and $y=6$
So, coordinates of B are $(3,6)$. Given that
$\mathrm{BD} \perp \mathrm{AB}$ and $\mathrm{CD}=3$ unit
$\mathrm{BC}=\sqrt{(2-3)^{2}+(4-6)^{2}}=\sqrt{1+4}=\sqrt{5}$
In right angled $\Delta \mathrm{BCD}, \mathrm{CD}^{2}=\mathrm{BC}^{2}+\mathrm{BD}^{2}$
$\Rightarrow 9=5+\mathrm{BD}^{2} \Rightarrow \mathrm{BD}^{2}=4 \Rightarrow \mathrm{BD}=2 \mathrm{unit}$
$\frac{1+x}{2}=2$ and $\frac{2+y}{2}=4$
$\Rightarrow \quad x=3$ and $y=6$
So, coordinates of B are $(3,6)$. Given that
$\mathrm{BD} \perp \mathrm{AB}$ and $\mathrm{CD}=3$ unit
$\mathrm{BC}=\sqrt{(2-3)^{2}+(4-6)^{2}}=\sqrt{1+4}=\sqrt{5}$
In right angled $\Delta \mathrm{BCD}, \mathrm{CD}^{2}=\mathrm{BC}^{2}+\mathrm{BD}^{2}$
$\Rightarrow 9=5+\mathrm{BD}^{2} \Rightarrow \mathrm{BD}^{2}=4 \Rightarrow \mathrm{BD}=2 \mathrm{unit}$
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