Question

A line is such that its segments between the straight lines 5x – y = 4 and 3x + 4y – 4 = 0 is bisected at the points (1, 5). Its equation is

• A. 23x – 7y + 6 = 0
• B. 7x + 4y + 3 = 0
• C. 83x – 35y + 92 = 0
• D. None of these

Answer 1

The correct option is C 83x – 35y + 92 = 0

(1, 5) is the mid-point of AB where A and B are on 1st and 2nd line but in opposite directions.
$\therefore \frac{x-1}{cos\theta }=\frac{y-5}{sin\theta }=\frac{r}{forA}=\frac{-r}{forB}...........\left(i\right)$
A lies on 1st,
$5(rcos\theta +1)–(rsin\theta +5)=4$
B lies on 2nd,
$3(–rcos\theta +1)+4(–rsin\theta +5)=4$
$\therefore r(5cos\theta –sin\theta )=4andr(3cos\theta +4sin\theta )=19$
$\therefore tan\theta =\frac{83}{35}$
$\therefore$ Equation of bisector is 83x – 35y + 92 = 0