Straight Lines Question 13
Question 13 - 2024 (31 Jan Shift 2)
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
(1) $\frac{15 \sqrt{5}}{7}$
(2) $\frac{17 \sqrt{ } 5}{6}$
(3) $\frac{17 \sqrt{ } 5}{7}$
(4) $\frac{\sqrt{5}}{17}$
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Answer (3)
Solution
$\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)$
C $\frac{2: 1}{A}$
$(-6,-8) \quad(a, b) \quad(3,4)$
$\Rightarrow a=0, b=0 \quad \Rightarrow P(3,5)$
Distance from $P$ measured along $x-2 y-1=0$ $\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta$
Where $\tan \theta=\frac{1}{2}$
$r(2 \cos \theta+3 \sin \theta)=-17$
$\Rightarrow r=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7}$
