Mathematics · Straight lines

JEE Main 2024 — 8 April, Shift 1 — Question 22

If the orthocentre of the triangle formed by the lines $2 x+3 y-1=0, x+2 y-1=0$ and $\mathrm{ax}+\mathrm{by}-1=0$ ,

is the centroid of another triangle, whose circumecentre and orthocentre respectively are $(3,4)$ and $(-6,-8)$ ,

then the value of $|a-b|$ is

Answer: 16

Numerical answer — enter this value.

$2 \mathrm{x}+3 \mathrm{y}-1=0$

$x+2 y-1=0$

$a x+b y-1=0$

$\left( \frac{6-6}{3},\frac{8-8}{3} \right)$ $=\left( 0,0 \right)$

$a x+b y-1=0$

$\left(\frac{1-0}{-1-0}\right)\left(\frac{-a}{b}\right)=-1$

$\Rightarrow-\mathrm{a}=\mathrm{b}$

$\Rightarrow \quad \mathrm{ax}-\mathrm{ay}-1=0$

$\mathrm{ax}-\mathrm{a}\left(1-\frac{2 \mathrm{x}}{3}\right)-1$

$x\left(a+\frac{2 a}{3}\right)=\frac{a}{3}$

$x=\frac{a+3}{5 a}$

$2\left(\frac{a+3}{5 a}\right)+3 y-1=0$

$y=\frac{1-\frac{2 a+6}{5 a}}{3}=\frac{3 a-6}{3 \times 5 a}$

$y=\frac{a-2}{5 a}$

$\frac{\left(\frac{a-2}{5 a}\right)}{\left(\frac{a+3}{5 a}\right)}=2 \Rightarrow a-2=2 a+6$

$\mathrm{a}=-8$ $\mathrm{b}=8$

$-8 x+8 y-1=0$

$|a-b|=16$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image