Question

If the orthocentre of the triangle formed by the lines \(2 x+3 y-1=0, x+2 y-1=0\) and ax + 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 1

Correct answer is : 16 

\(2 x+3 y-1=0\)

\(\mathrm{x}+2 \mathrm{y}-1=0\)

ax + by \( -1=0\)

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

\(=(0,0)\)  

  

ax + by - 1 = 0

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

\(\Rightarrow-\mathrm{a}=\mathrm{b}\)

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

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

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

\(x=\frac{a+3}{5 a}\)

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

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

\(y=\frac{a-2}{5 a}\)

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

\(\mathrm{a}=-8\)

\(\mathrm{b}=8\)

\(-8 x+8 y-1=0\)

\(|a-b|=16\)