Question

Solution

Correct option is

11x + 3y + 17 = 0

Write the equation so that constant terms are positive.

–3x – 4y + 11 = 0    and   –12x + 5y + 2 = 0

a1a2 + b1b2 = (–3) ( –12) + (–4) (5) = 16 (positive)

Hence +ve sign give the obtuse angle bisector.

⇒         –ve sign gives the acute angle bisector.

⇒         acute bisector is given as: ⇒        13(–3x – 4y + 11) = –5(–12x + 5y + 2)

⇒        11x + 3y + 17 = 0 is the acute angle bisector.

SIMILAR QUESTIONS

Q1

A and B are two fixed points. The vertex C of a Δ ABC moves such that cot A + cot B = constant. Locus of C is a straight line:

Q2

The number of integer values of m, for which the x-co-ordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is:

Q3

The vertices of a triangles are A (x1x1 tan α), B (x2x2 tan β) and C (x3,x3 tan γ). If the circumcentre of Δ ABC coincides with the origin and H (ab) be its orthocenter, then a/b is equal to:

Q4

A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinates axes at points P and Q. As L varies, the absolute minimum values of OP + OQ is (O is origin)

Q5

Consider the family of lines

(x + y – 1) + λ (2x + 3y – 5) = 0

and   (3x + 2y – 4) + µ (x + 2y – 6) = 0

equation of a straight line that belongs to both the families is:

Q6

If p1p2p3 be the length perpendicular from the points

(m2, 2m), (mm’m + m’) and (m’2, 2m’)

respectively on the line Q7

The point (a2a + 1) is a point in the angle between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0 containing origin. Then ‘a’ belongs to the interval.

Q8

Find all points on x + y = 4 that lie at a unit distance from the line 4x + 3y– 10 = 0.

Q9

Find the equation of the obtuse angle bisector of the lines 12x – 5y + 7 = 0 and 3y – 4x – 1 = 0.

Q10

Find the equation of the locus of a moving point so that its distance from the point (1, 0) is always twice the distance from the point (0, –2).