The equation of straight line passing through the point of intersection of family of lines x(3+4 λ)+y(4+3 λ)+1+6 λ=0 which is at a minimum distance from the point (1,1) can be expressed as x+b y-c=0, where b and c are natural numbers, then the value of (b+c) is

Question

The equation of straight line passing through the point of intersection of family of lines x(3+4 λ)+y(4+3 λ)+1+6 λ=0 which is at a minimum distance from the point (1,1) can be expressed as x+b y-c=0, where b and c are natural numbers, then the value of (b+c) is (A) 5 (B) 8 (C) 9 (D) 10

Tardigrade Admin · Accepted Answer

Family of line is (3 x+4 y+1)+λ(4 x+3 y+6)=0 Point of intersection is (-3,2) Line at minimum distance from (1,1) and passing through (-3,2) is y-2=-(1/4)(x+3) x+4 y-5=0 ⇒ b=4, c=5

Q. The equation of straight line passing through the point of intersection of family of lines x(3+4λ)+y(4+3λ)+1+6λ=0 which is at a minimum distance from the point (1,1) can be expressed as x+by−c=0, where b and c are natural numbers, then the value of (b+c) is

5

8

9

10

Family of line is (3x+4y+1)+λ(4x+3y+6)=0 Point of intersection is (−3,2) Line at minimum distance from (1,1) and passing through (−3,2) is y−2=−41​(x+3) x+4y−5=0 ⇒b=4,c=5

Family of line is $(3 x + 4 y + 1) + λ (4 x + 3 y + 6) = 0$
Point of intersection is $(- 3, 2)$ Line at minimum distance from $(1, 1)$ and passing through
$(- 3, 2)$ is $y - 2 = - \frac{1}{4} (x + 3)$
$x + 4 y - 5 = 0$
$\Rightarrow b = 4, c = 5$