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  4. 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

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Q. The equation of straight line passing through the point of intersection of family of lines $x(3+4 \lambda)+y(4+3 \lambda)+1+6 \lambda=0$ 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

Straight Lines

Solution:

Family of line is $(3 x+4 y+1)+\lambda(4 x+3 y+6)=0$
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 $
$\Rightarrow b=4, c=5$