Let L be the line belonging to the family of the straight lines (a + 2b) x + (a – 3b) y +a - 86 = 0,
a, b belongs to R, which is farthest from point P (2,2)
Q: The equation of line L is
Answers
Given (a+2b)x+(a−3b)y+a−8b=0
a(x+y+1)+b(2x−3y−8)=0
L
1
:x+y+1=0
L
2
:2x−3y−85=0
L
2
−2L
1
⇒(2x−3y−8=0)−(2x+2y+2=0)=−5y−10=0
y=−2,x=1
Any line belong to family of lines is λ(x+y+1)+(2x−3y−8)=0
x(2+λ)+y(a−3)+λ−8=0
distance from (2,2) should be maximum.
d
2
=(x−2)
2
+(y−2)
2
=(x−2)
2
+(
λ−3
(2+λ)x+(λ−8)
)
2
Farthest line is the line which give maximum perpendicular distance.
d=
(λ+2)
2
+(λ−3)
2
2(2+λ)+2(λ−3)+λ−8
=
2λ
2
+13−2λ
5λ−10
d
2
=(
2λ
2
+13−2λ
25(λ−2)
2
)
d
2
=(
2λ
2
+13−2λ
25(λ−2)
2
)
dλ
d(d
2
)
=2(λ−2)(2λ
2
−2λ+13)+(4λ−2)(λ−2)
2
=0
=2(2λ
2
−2λ+13)+(λ−2)(2λ−1)=0
(λ−2)(2λ
2
−2λ+13+2λ
2
−λ−4λ+2)=0
(λ−2)(2λ
2
−7λ+15)=0
⇒λ−2=0
⇒4x−y−7=0
a=
4
7
,b=
1
7
Δ=
2
1
ab=
4
49
×
2
1
=
8
49