Question

The equation of the line passing the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y -1 = 0 and whose distance from the point (2,-1) is 2 is

Answer 1

Line passing through point of intersecting of $L_1,L_2$ is given by $L_1+{\lambda }_2=0$

$\Rightarrow (3x+2y+4)+\lambda (2x+5y-1)=0$

$\Rightarrow x(3+2\lambda )+y(2+5\lambda )+4-\lambda =0$

Distance of $(2,-1)$ to this line,

$\left|\frac{2(3+2\lambda )-(2+5\lambda )+4-\lambda }{\sqrt{(3+2\lambda {)}^2+(2+5\lambda {)}^2}}\right|=2$

$\Rightarrow 6+4\lambda -2-5\lambda =2\sqrt{29{\lambda }^2+32\lambda +13}$

$\Rightarrow -2\lambda +8=2\sqrt{29{\lambda }^2+32\lambda +13}$

$\Rightarrow {\lambda }^2+16-8d=29{\lambda }^2+32\lambda +13$

$\Rightarrow 28{\lambda }^2+40\lambda -3=0\Rightarrow 28{\lambda }^2+42\lambda -2\lambda -3=0$

$14\lambda (2\lambda +3)-(2\lambda +3)=0$

$(14\lambda -1)(2\lambda +3)=0\Rightarrow \lambda =\frac{1}{14}$ or $\lambda =\frac{-3}{2}$

$L:x(3+\frac{1}{7})+y(2+\frac{5}{14})+4-\frac{1}{14}=0$

$44x+33y+55=0\Rightarrow 4x+3y+5=0$

or $x(3-3)+y(2-\frac{15}{2})+4+\frac{3}{2}=0$

$\Rightarrow y(-11)+11=0\Rightarrow y=1$