﻿ If θ is an angle between the lines given by the equation 6x2 + 5xy – 4y2 + 7x + 13y – 3 = 0, then equation of the line passing through the point of intersection of these lines and making an angle θ with the positive x-axis is  : Kaysons Education

# If θ Is An Angle Between The Lines Given By The Equation 6x2 + 5xy – 4y2 + 7x + 13y – 3 = 0, Then Equation Of The Line Passing Through The Point Of Intersection Of These Lines And Making An Angle θ With The Positive x-axis Is

#### Video lectures

Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.

#### Online Support

Practice over 30000+ questions starting from basic level to JEE advance level.

#### National Mock Tests

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

#### Organized Learning

Proper planning to complete syllabus is the key to get a decent rank in JEE.

#### Test Series/Daily assignments

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

## Question

### Solution

Correct option is

11x – 2y + 13 = 0

Writing the given equation as a quadratic in x we have

6x2 + (5y + 7)x – (4y2 – 13y + 3) = 0

⇒ 2x – y + 3 = 0 and 3x + 4y – 1 = 0

Which are the two lines represented by the given equation and the point of intersection is (–1, 1), obtained by solving these equations.

[The point of intersection of the given lives is also obtained by solving 12x + 5y + 7 = 0 and 5x – 8y + 13 = 0 (see Text, 15.4(4))]

So the equation of the required line is

#### SIMILAR QUESTIONS

Q1

On the portion of the straight line x + y = 2 which is intercepted between the axes, a square is constructed, away from the origin, with this portion as one of its side. If p denotes the perpendicular distance of a side of this square from the origin, then the maximum value of p is

Q2

The line x + y = 1 meets x-axis at A and y-axis at BP is the mid-point ofAB (Fig). P is the foot of the perpendicular from P to OAM1 is that from P1 to OPP2 is that from M1 to OAM2 is that from P2 to OPP3is that from M2 to OA and so on. If Pn denotes the nth foot of the perpendicular on OA from Mn – 1, then OPn =

Q3

The line x + y = a, meets the axis of x and y at A and B respectively. A triangle AMN is inscribed in the triangle OABO being the origin, with right angle at NM and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to.

Q4

The point (4, 1) undergoes the following transformation successively.

(i) Reflection about the line y = x

(ii) Translation through a distance 2 units along the positive direction of x-axis.

(iii) Rotation through an anlge π/4 about the origin in the anticlockwise direction

(iv) Reflection about x = 0

The final position of the given point is

Q5

A line cuts the x-axis at (7, 0) and the y-axis at B(0, –5). A variable linePQ is draw perpendicular to AB cutting the x-axis at P and the y-axis at in θ. If AQ and BP intersect at R, the locus of R is

Q6

Equation of a line which is parallel to the line common to the pair of lines given by 6x2 – xy – 12y2 = 0 and 15x2 + 14xy – 8y2 = 0 and the sum of whose intercepts on the axes is 7, is

Q7

If pairs of lines x2 + 2xy + ay2 = 0 and ax2 + 2xy + y2 = 0 have exactly one line in common then the joint equation of the other two lines is given by

Q8

If the lines joining the origin to the intersection of the line y = mx + 2 and the curve x2 + y2 = 1 are at right angles, then

Q9

Let PQR be a right angled isosceles triangle right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is

Q10

If one of the lines given by the equation 2x2 + axy + 3y2 = 0 coincide with one of those given by 2x2 + bxy – 3y2 = 0 and the other lines represented by them be perpendicular, then