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 \[\theta\] 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

\[\frac{x - x_1}{cos\theta} = \frac{y - y_1}{sin\theta}\]

\[ \Rightarrow \frac{x - 1}{cos\theta} = \frac{y - 5}{sin\theta}\]

\[ \Rightarrow \frac{y - 5}{x - 1} = \tan\theta\]

Let \[A P_1 = A P_2 = r\]

Then, the coordinates of \[P_1 \text { and } P_2\] are given by \[\frac{x - 1}{cos\theta} = \frac{y - 5}{sin\theta} = r \text { and } \frac{x - 1}{cos\theta} = \frac{y - 5}{sin\theta} = - r\]

 So, the coordinates of \[P_1 \text { and } P_2\] are  \[\left( 1 + rcos\theta, 5 + r\sin\theta \right) \text { and } \left( 1 - rcos\theta, 5 - r\sin\theta \right)\] respectively.
Clearly,

\[P_1 \text { and } P_2\] lie on 5x − y − 4 = 0 and 3x + 4y − 4 = 0, respectively.

\[\therefore 5\left( 1 + rcos\theta \right) - 5 - r\sin\theta - 4 = 0 \text { and } 3\left( 1 - rcos\theta \right) + 4\left( 5 - r\sin\theta \right) - 4 = 0\]

\[ \Rightarrow r = \frac{4}{5cos\theta - sin\theta} \text { and } r = \frac{19}{3cos\theta + 4sin\theta}\]

\[ \Rightarrow \frac{4}{5cos\theta - sin\theta} = \frac{19}{3cos\theta + 4sin\theta}\]

\[ \Rightarrow 95cos\theta - 19sin\theta = 12cos\theta + 16sin\theta\]

\[ \Rightarrow 83cos\theta = 35sin\theta\]

\[ \Rightarrow tan\theta = \frac{83}{35}\]

Thus, the equation of the required line is

\[\frac{y - 5}{x - 1} = tan\theta\]

\[ \Rightarrow \frac{y - 5}{x - 1} = \frac{83}{35}\]

\[ \Rightarrow 83x - 35y + 92 = 0\]