The equation of the straight line in the norm | vedclass

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(B) Let the required line be $x+2y+k=0$. Since it divides the distance between $x+2y+3=0$ and $x+2y+8=0$ in the ratio $1:2$ internally,the constant $k

Vedclass · Accepted Answer

$x \cos \alpha+y \sin \alpha=\frac{14}{\sqrt{45}}, \alpha=\pi+	an ^{-1} 2$

Vedclass · Answer

(B) Let the required line be $x+2y+k=0$. Since it divides the distance between $x+2y+3=0$ and $x+2y+8=0$ in the ratio $1:2$ internally,the constant $k$ satisfies $\frac{|k-3|}{|k-8|} = \frac{1}{2}$. $2|k-3| = |k-8| \Rightarrow 4(k^2-6k+9) = k^2-16k+64$. $3k^2-8k-28=0 \Rightarrow (3k-14)(k+2)=0$. Since the line lies between the two given lines,$k$ must be between $3$ and $8$,so $k = \frac{14}{3}$. The equation is $x+2y+\frac{14}{3}=0$,or $3x+6y+14=0$. To convert to normal form $x \cos \alpha + y \sin \alpha = p$,we write $-3x-6y=14$. Dividing by $\sqrt{(-3)^2+(-6)^2} = \sqrt{45}$,we get $\frac{-3}{\sqrt{45}}x + \frac{-6}{\sqrt{45}}y = \frac{14}{\sqrt{45}}$. Here $\cos \alpha = \frac{-3}{\sqrt{45}} = \frac{-1}{\sqrt{5}}$ and $\sin \alpha = \frac{-6}{\sqrt{45}} = \frac{-2}{\sqrt{5}}$. Since both $\cos \alpha$ and $\sin \alpha$ are negative,$\alpha$ lies in the third quadrant. $\alpha = \pi + 	an^{-1}(\frac{-2/-1}) = \pi + 	an^{-1}(2)$.

The equation of the straight line in the normal form which is parallel to the lines $x+2y+3=0$ and $x+2y+8=0$ and divides the distance between these two lines in the ratio $1:2$ internally is

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