Question

A line is such that its segment between the straight lines 5x − y − 4 = 0 and 3x + 4y − 4 = 0 is bisected at the point (1, 5). Obtain its equation.

Answer 1

Let P₁P₂ be the intercept between the lines 5x − y − 4 = 0 and 3x + 4y − 4 = 0.
Let P₁P₂ make an angle $\mathrm{\theta }$ with the positive x-axis.

Here, $\left(x_1,y_1\right)=A\left(1,5\right)$

So, the equation of the line passing through A (1, 5) is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x-1}{{\displaystyle \mathrm{cos\theta }}}=\frac{y-5}{{\displaystyle \mathrm{sin\theta }}} \\ \Rightarrow \frac{y-5}{x-1{\displaystyle }}=\tan \theta \end{gathered}$$




Let $A{P}_1=A{P}_2=r$
Then, the coordinates of $P_1\mathrm{and}{P}_2$ are given by

$\frac{x-1}{{\displaystyle \mathrm{cos\theta }}}=\frac{y-5}{{\displaystyle \mathrm{sin\theta }}}=r\mathrm{and}\frac{x-1}{{\displaystyle \mathrm{cos\theta }}}=\frac{y-5}{{\displaystyle \mathrm{sin\theta }}}=-r$

So, the coordinates of $P_1\mathrm{and}{P}_2$ are $\left(1+r\mathrm{cos\theta },5+r\sin \theta \right)\mathrm{and}\left(1-r\mathrm{cos\theta },5-r\sin \theta \right)$, respectively.

Clearly, $P_1\mathrm{and}{P}_2$ lie on 5x − y − 4 = 0 and 3x + 4y − 4 = 0, respectively.

$$\begin{gathered} \therefore 5\left(1+r\mathrm{cos\theta }\right)-5-r\sin \theta -4=0\mathrm{and}3\left(1-r\mathrm{cos\theta }\right)+4\left(5-r\sin \theta \right)-4=0 \\ \Rightarrow r=\frac{4}{5\mathrm{cos\theta }-\mathrm{sin\theta }}\mathrm{and}r=\frac{19}{3\mathrm{cos\theta }+4\mathrm{sin\theta }} \\ \Rightarrow \frac{4}{5\mathrm{cos\theta }-\mathrm{sin\theta }}=\frac{19}{3\mathrm{cos\theta }+4\mathrm{sin\theta }} \\ \Rightarrow 95\mathrm{cos\theta }-19\mathrm{sin\theta }=12\mathrm{cos\theta }+16\mathrm{sin\theta } \\ \Rightarrow 83\mathrm{cos\theta }=35\mathrm{sin\theta } \\ \Rightarrow \mathrm{tan\theta }=\frac{83}{35} \end{gathered}$$

Thus, the equation of the required line is

$$\begin{gathered} \frac{y-5}{x-1}=\mathrm{tan\theta } \\ \Rightarrow \frac{y-5}{x-1}=\frac{83}{35} \\ \Rightarrow 83x-35y+92=0 \end{gathered}$$