AP EAMCET Engineering 2015 Code C Paper

Let the formed triangle be ABC.
Family of straight lines is
AB+λAC→2x+3y−1+λ(x+2y−1)=0
It represents altitude through A
So the given line passes through ortho centre and we get λ=−1
So the given line AO is x+y=0
AO is perpendicular to BC
[Using the concept that if line L1 and L2 are at right angles to each other then
m1×m2=−1
where m1 and m2 are slopes of L1 and L2 respectively ]
(−1)×‌

−a

b

=−1
∴a=−b
BA+µBC=0 as it passes through (0,0) we get µ=−1
This line is perpendicular to AC
∴−‌

2−a

3+a

×(−‌

1

2

)=−1
⇒2−a=−6−2a
∴a=−8‌ and ‌b=8