TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

Equation of line passing through the point of intersection of lines x+2y−19=0 and x−2y−3=0 is
(x+2y−19)+λ(x−2y−3)=0
is at a
perpendicular distance of 5 units from point (−2,4)
So, 5=‌

|−2−2λ+8−8λ−19−3λ|

√(1+λ)2+4(1−λ)2

⇒‌‌25[5λ2−6λ+5]=(13λ+13)2
⇒‌‌44λ2+488λ+44=0
⇒λ=−11 or −‌

1

11

∴ The possible equation are
Now, on comparing with 5x+by+c=0, we get The possible value of 5+b+c=−14 or −86