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The distance of the origin from the cetroid of the triangle whose two sides have the equations $\mathrm{x-2 y+1=0 \text { and } 2 x-y-1=0}$ and whose orthocentre is $\mathrm{\left(\frac{7}{3}, \frac{7}{3}\right)}$ is:
Option: 1
$\sqrt{2}$

Option: 2
$2$

Option: 3
$2\sqrt{2}$

Option: 4
$4$

Answers (1)

best_answer

Altitude $\mathrm{BD}$ is perpendicular to $\mathrm{A C \quad(x-2 y+1=0)}\\$

$\mathrm{\therefore\: Let \: B D\: is \: 2 x+y+\lambda=0}$

It passes through $\mathrm{D}$

$\mathrm{\therefore \frac{14}{3}+\frac{7}{3}+\lambda=0} \\$

$\mathrm{\Rightarrow \lambda=-7 }\\$

$\mathrm{\therefore \text { BD is } 2 x+y-7=0}$

Solving it with equation of $\mathrm{AB}$ we get $\mathrm{B(2,3)}$

Similarly we can get coordinates of $\mathrm{C}$ as $\mathrm{(3,2)}$

Centroid of $\mathrm{\Delta ABC}$

$\mathrm{=\left(\frac{1+2+3}{3}, \frac{1+3+2}{3}\right)} \\$

$\mathrm{=(2,2)}$

Distance of centroid from origin

$\mathrm{=\sqrt{4+4}=2 \sqrt{2}}$

Hence answer is option 3

Posted by

Shailly goel

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