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Question : 43 of 160
Marks:
+1 ,
-0
Solution:
A ( − 3 , 4 ) , C ( 0 , 1 ) Let
R ≡ ( x , y )
‌ A B + B C = ‌ minimum ‌ ‌ √ ( x + 3 ) 2 + ( y − 4 ) 2 + √ x 2 + ( y − 1 ) 2 is minimum
B is on
2 x + y = 7 ⇒ y = 7 − 2 x √ ( x + 3 ) 2 + ( 7 − 2 x − 4 ) 2 + √ x 2 + ( 7 − 2 x − 1 ) 2 . . . . . . ( i i )
⇒ √ x 2 + 9 + 6 x + 9 + 4 x 2 − 12 x ‌ ‌ minimum
+ √ x 2 + 36 + 4 x 2 − 24 x ‌ ‌ minimum
⇒ √ 5 x 2 − 6 x + 18 + √ 5 x 2 − 24 x + 36 minimum
On differentiating w.r.t.
x ,
‌ ⇒ ( 10 x − 6 ) √ 5 x 2 − 24 x + 36 ‌ = − ( 10 x − 24 ) ⋅ √ 5 x 2 − 6 x + 18 ⇒ ‌ ‌ ( 10 x − 6 ) 2 ( 5 x 2 − 24 x + 36 ) = ‌ ( 10 x − 24 ) 2 ( 5 x 2 − 6 x + 18 ) ⇒ ‌ ‌ ( 25 x 2 + 9 − 30 x ) ( 5 x 2 − 24 x + 36 ) = ‌ ( 25 x 2 + 144 − 120 x ) ( 5 x 2 − 6 x + 18 ) ⇒ 25 x 2 − 192 x + 252 = 0 ⇒ 25 x 2 − 186 x − 6 x + 252 = 0 ‌ ⇒ ‌ ‌ 25 x 2 − 186 x − 6 x + 252 = 0 ‌ ⇒ ‌ ‌ x = 6 or
x = ‌ = ‌ ‌ ∵ ‌ ‌ 2 x + y = 7 For
x = 6 ⇒ 2 ( 6 ) + y = 7 ⇒ y = − 5 For
x = ‌ ⇒ 2 ( ‌ ) + y = 7 ⇒ ‌ + y = 7 ⇒ y = ‌ When we substitute the value
( x , y ) = ( 6 , − 5 ) and
( ‌ , ‌ ) in Eq. (i) or when we substitute
x = 6 and
x = ‌ in Eq. (ii), we get
Minimum value occurs when
x = 42 / 25 ∴ ‌ ‌ B ‌ ≡ ( ‌ , ‌ ) and
A B ‌ = √ ( ‌ + 3 ) 2 + ( ‌ − 4 ) 2 ‌ = √ ( ‌ ) 2 + ( ‌ ) 2 = √ ‌
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