Question

For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, the equation of the bisector of the angle which contains the origin is

• A. 7x + 9y + 3 = 0
• B. 7x – 9y + 3 = 0
• C. 7x + 9y – 3 = 0
• D. None of these

Answer 1

The correct option is C

7x + 9y – 3 = 0

Re-writing the given equations so that constant terms are positive, we have

–4x – 3y + 6 = 0 …… (1)

and 5x + 12y + 9 = 0 …… (2)

Now, $a_1{a}_2+b_1{b}_2$ = (–4)(5) + (–3)(12)

= –20 – 36 = –56 < 0

So, the origin lies in acute angle.

$\therefore$ The equation of the bisector of the acute angle between the lines (1) and (2) is

$\frac{-4x-3y+6}{\sqrt{(-4{)}^2+(-3{)}^2}}=+\frac{5x+12y+9}{\sqrt{5^2+(12{)}^2}}$

$\Rightarrow$ –52x – 39y + 78 = 25x + 60y + 45

$\Rightarrow$ 7x + 9y – 3 = 0