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 ⇒‌‌(1+λ)x+(2−2λ)y−(19+3λ)=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 ⇒125λ2−150λ+125=169λ2+338λ+169
⇒‌‌44λ2+488λ+44=0 ⇒‌‌11λ2+122λ+11=0⇒(11λ+1)(λ+11)=0
⇒λ=−11 or
−‌ ∴ The possible equation are
−10x+24y+14=0⇒5x−12y−7=0
or 10x+24y−206=0⇒5x+12y−103=0
Now, on comparing with
5x+by+c=0, we get The possible value of
5+b+c=−14 or
−86