## Question

### Solution

Correct option is

6x + 2y – 5 = 0

Firstly, make the constant terms (c1c2) positive then

–x – y + 3 = 0 and 7x – y + 5 = 0 = –7 + 1 = –6

i.e.,   a1a2 + b1b2 < 0

Hence “+” sign gives the acute bisector.   ∴   12x + 4y – 10 = 0   or   6x + 2y – 5 = 0

is the acute angle bisector.

#### SIMILAR QUESTIONS

Q1

Find the equation of the line passing through the point of intersection of the lines

x + 5y + 7 = 0, 3x + 2y – 5 = 0 and

1. parallel to the line 7x + 2y – 5 = 0

2. perpendicular to the line 7x + 2y – 5 = 0

Q2

Find the equation of straight line which passes through the intersection of the straight lines

3x – 4y + 1 = 0 and 5x + y – 1 = 0

and cuts off equal intercepts from the axes.

Q3

Find the orthocentre of the triangle of the triangle ABC whose angular points are A(1, 2), B(2, 3) and C(4, 3).

Q4

If the orthocentre of the triangle formed by the lines

2x + 3y – 1 = 0, x + 2y – 1 = 0, ax + by – 1 = 0   is at origin, then find (a,b).

Q5

Find the equations of the straight lines passing through the point (2, 3) and inclined at π/4 radians to the line 2x + 3y = 5.

Q6

Find the equations to the straight lines passing through the point (2, 3) and equally inclined to the lines 3x – 4y – 7 = 0 and

12x – 5y + 6 = 0

Q7

Find the equations of angular bisector bisecting the angle containing the origin and not containing the origin of the lines

4x + 3y – 6 = 0 and 5x + 12y + 9 = 0.

Q8

Find the equation of the bisector of the obtuse angle between the lines 3x– 4y + 7 = 0 and 12x + 5y – 2 = 0.