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QJEEM24Ma77ed: JEE Main 2024 Mathematics - Class 11 JEEM24Ma77ed Solution: Let and respectively denote the centroid, circumcentre and…

Let A(a,b),B(3,4)A(a, b), B(3,4) and C(−6,−8)C(-6,-8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a+3,7b+5)P(2 a+3,7 b+5) from the line 2x+3y−4=02 x+3 y-4=0 measured parallel to the line x−2y−1=0x-2 y-1=0 is

Multiple Choice Options

  • A 1756\frac{17 \sqrt{5}}{6}
  • B 1557\frac{15 \sqrt{5}}{7}
  • C 1757\frac{17 \sqrt{5}}{7}
  • D 517\frac{\sqrt{5}}{17}

Step-by-Step Verified Solution

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

JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 28 English Explanation

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

Distance from P\mathrm{P} measured along x−2y−1=0\mathrm{x}-2 \mathrm{y}-1=0

⇒x=3+rcos⁡θ,y=5+rsin⁡θ\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta

amp; Where tan⁡θ=12amp;r(2cos⁡θ+3sin⁡θ)=−17amp;⇒r=∣−1757∣=1757\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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