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(2a+3,7b+5) from the line 2x+3y−4=0 measured parallel to the line x−2y−1=0 is
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Official 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 \[\mathrm{P}\] measured along \[\mathrm{x}-2 \mathrm{y}-1=0\]
\[\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta\]
\[ Where tanθ=21r(2cosθ+3sinθ)=−17⇒r=7−175=7175\]