The distance between the lines represented by $4 x^2+20 x y+25 y^2+2 x+5 y-12=0$ is equal to

Question

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } 4 x^2+(20 y+2) x+\left(25 y^2+5 y-12ight) \ & x=\frac{-(20 y+2) \pm \sqrt{(20 y+2)^2-4 	imes 4\left(25 y^2+5 y-12ight)}}{2 	imes 4} \ & \Rightarrow x=-20 y-2 \pm \sqrt{\frac{400 y^2+4+80 y-400 y^2}{-80 y+192}} \ & \Rightarrow x=\frac{-20 y-2 \pm 14}{8} \Rightarrow 8 x+20 y=-2 \pm 14 \ & \Rightarrow 8 x+20 y-12=0 	ext { and } 8 x+2 y+16=0 \end{aligned} $ The given two lines are parallel $ \begin{aligned} d & =\frac{|16+12|}{\sqrt{8^2+(20)^2}} \ & =\frac{28}{\sqrt{64+400}}=\frac{28}{\sqrt{464}}=\frac{7}{\sqrt{29}} \end{aligned} $

The distance between the lines represented by $4 x^2+20 x y+25 y^2+2 x+5 y-12=0$ is equal to

Solution