AB:2x+3y−1=0 AC:x+2y+1=0 BC:ax+by−1=0 AD:(2x+3y−1)+λ(x+2y+1)=0 AD passes through origin ⇒−1+λ=0 or λ = 1 ∴ AD,3x+5y=0 ∵ AD⊥BC ⇒(
−3
5
)(
−a
b
)=−1‌‌‌‌‌[m1.m1=−1] ⇒3a+5b=0 ...(i) Again, BE,(2x+3y−1)+µ(ax+by−1)=0 BE passes through origin ⇒−1−µ=0 ⇒µ=−1 BE,(2−a)x+(3−b)y=0 ∵ BE ⊥ AC ⇒(
−1
2
)−(
2−a
3−b
)=−1 ⇒a+2b=8 ...(ii) Soving Eqs. (i) and (ii), we get a=−40,b=24 ∴