Question

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

Answer 1

$\text{Let}{P}_{1}P2\text{be the intercept between the lines}5x-y-4=0and3x+4y-4=0$

$\text{Let}{P}_{1}P2\text{make an angle}\theta \text{with the positive x-axis}$

$\text{Here,}(x_1,y_1)=A(1,5)$

$\text{So, the equation of the line passing through}A(1,5)is$

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

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

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

$LetA{P}_1=A{P}_2=r$

$\text{Then, the coordinates of}{P}_{1}and{P}_2\text{are given by}$

$\frac{x-1}{cos\theta }=\frac{y-5}{sin\theta }=rand\frac{x-1}{cos\theta }=\frac{y-5}{sin\theta }$

$=-r$

$\text{So, the coordinates of}{P}_{1}and{P}_{2}are(1+rcos\theta ,5+rsin\theta )and(1-r.cos\theta ,5-rsin\theta ),\text{respectively.}$

$\text{Clearly,}{P}_{1}and{P}_2\text{lie on}5x-y-4=0$

$and3x+4y-4=0,\text{respectively.}$