JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 19)
Let $$A(a, b), B(3,4)$$ and $$C(-6,-8)$$ 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
$$\frac{17 \sqrt{5}}{6}$$
$$\frac{15 \sqrt{5}}{7}$$
$$\frac{17 \sqrt{5}}{7}$$
$$\frac{\sqrt{5}}{17}$$
Explanation
$$\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)$$
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$$\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)$$
Distance from $$\mathrm{P}$$ measured along $$\mathrm{x}-2 \mathrm{y}-1=0$$
$$\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta$$
$$\begin{aligned} & \text { Where } \tan \theta=\frac{1}{2} \\ & \mathrm{r}(2 \cos \theta+3 \sin \theta)=-17 \\ & \Rightarrow \mathrm{r}=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7} \end{aligned}$$
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