Let A(a, b), B(3,4) and (-6,-8) respectively denote centroid, circumcentre and orthocentre of a tria

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9.Straight Line

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Let $A(a, b), B(3,4)$ and $(-6,-8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P(2 a+3,7 b+5)$ from the line $2 x+3 y-4=0$ measured parallel to the line $x-2 y-1=0$ is

A

$\frac{15 \sqrt{5}}{7}$

B

$\frac{17 \sqrt{5}}{6}$

C

$\frac{17 \sqrt{5}}{7}$

D

$\frac{\sqrt{5}}{17}$

(JEE MAIN-2024)

Solution

$\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)$

$\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)$

Distance from $P$ measured along $x-2 y-1=0$

$\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta$

Where $ \tan \theta=\frac{1}{2} $

$ r(2 \cos \theta+3 \sin \theta)=-17 $

$ \Rightarrow r=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7}$

Standard 11

Mathematics

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