Find the equations of the line passing through the point $(2,3)$ and making intercept of length 2 un

Question

Find the equations of the line passing through the point $(2,3)$ and making intercept of length 2 units between the lines $y+2 x=3$ and $y+2 x=5$.

Professor PC Thomas Classes · Answer

The equation of the line through a given point $(2,3)$ is $\frac{x-2}{\cos 	heta}=\frac{y-3}{\sin 	heta}=r$\ $\Rightarrow x=2+r \cos 	heta$ and $y=3+r \sin 	heta$\ The given lines are: $$\begin{equation*}y+2 x=3 	ag{ii} \end{equation*}$$ $$\begin{equation*}y+2 x=5 	ag{iii} \end{equation*}$$ Now the line above makes an intercept of 2 between the given parallel lines, i.e. $\left|r_{1}-r_{2}ight|=2$\ Now putting the values of $(x, y)$ from equation (i) into equation (ii) after changing $r ightarrow r_{1}$ we have $3+r_{1} \sin 	heta+2\left(2+r_{1} \cos 	hetaight)=3$ Taking the value of $(x, y)$ from (i) change $r ightarrow r_{2}$ and put in equation (ii) we get $$\begin{equation*}3+r_{2} \sin 	heta+2\left(2+r_{2} \cos 	hetaight)=5 	ag{v} \end{equation*}$$ From (iv) we have\ $r_{1}=\frac{-4}{\sin 	heta+2 \cos 	heta}, r_{2}=\frac{-2}{\sin 	heta+2 \cos 	heta}$\ Now as give $\left|r_{1}-r_{2}ight|=2=\frac{2}{\sin 	heta+2 \cos 	heta}$ $$\begin{equation*}	herefore \sin 	heta+2 \cos 	heta=1 \Rightarrow \sin 	heta=1-2 \cos 	heta 	ag{vi} \end{equation*}$$ Putting this in formula $\sin ^{2} 	heta+\cos ^{2} 	heta=1$\ $\Rightarrow(1-2 \cos 	heta)^{2}+\cos ^{2} 	heta=1$\ $\Rightarrow 5 \cos ^{2} 	heta-4 \cos 	heta=0, \cos 	heta=0$ or $4 / 5$\ Putting this value in equation (vii) we get $\sin 	heta=1$ or $-3 / 5$ The lines are $x-2=0$ and $\frac{x-2}{4 / 5}=\frac{y-3}{-3 / 5}$\ or $\quad 3 x+4 y=-18$\ or $\quad 3 x+4 y=-18$\ $	herefore$ The lines are $x-2=0$ and $3 x+4 y-18=0$.\