# PAIR OF STRAIGHT LINES

### GENERAL EQUATION OF A PAIR OF STRAIGHT LINES

Consider the equations of two straight lines

Their combined equation may be given by

. ......(1)

On multiplying the left hand side of (1), we get an equation of the form ......(2)

where a = a1a2, 2h = a1b2 + a2b1, ...... etc.

The equation (2) is the most general equation of second degree, which will represent a pair of straight lines under certain conditions.

### HOMOGENEOUS EQUATIONS OF SECOND DEGREE

An equation of the type

......(3)

is a homogeneous equation of second degree. [Note that the sum of the powers of x and y in every term is the same and 2].

The homogeneous equation of the second degree always represents a pair of straight lines passing through the origin. We have

- If h2 > ab, the two lines are REAL AND DIFFERENT.
- If h2 = ab, the two lines are COINCIDENT.
- If h2 < ab, the two lines are IMAGINARY having origin as their real point of intersection.

If represents the pair of lines y = m1x and y = m2x. Then

........(4)

Comparing the coefficients we get

and ........(5)

### ANGLE BETWEEN THE LINES GIVEN BY ax2 + 2hxy + by2 = 0

If the equation ax2 + 2hxy + by2 = 0 represents the straight lines y = m1x and y = m2x, then the acute angle Î¸, between the lines is given by

### BISECTORS OF THE ANGLES BETWEEN THE LINES ax2 + 2hxy + by2 = 0

The equation of the pair of lines bisecting the angle between the lines given byis or

IMPORTANT RESULTS

- Two lines are coincident if tan Î¸ = 0 i.e. if h2 – ab = 0
- Two lines are perpendicular of tan Î¸ = ∞ i.e. if a + b = 0

∴ The equation of a pair of perpendicular straight lines can be given by ax2 + 2hxy – ay2 = 0

[Substituting b = –a]

or, where

- If the lines given by the equation are equally inclined to axes, then the coordinate axes are the bisectors, i.e. the equation of pair of bisector must be xy = 0. Therefore h = 0.
- If two pairs of straight lines are equally inclined to one another, then both must have the same pair of bisectors.
- Since coeff. of x2 + coeff. of y2 = 0. Therefore two bisectors are always perpendicular. (The original lines may or may not be perpendicular).

### CONDITION THAT THE GENERAL EQUATION OF SECOND DEGREE MAY REPRESENT A PAIR OF STRAIGHT LINES

Let the general equation of the second degree be

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ...(1)

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ...(1)

It will represent a pair of straight lines if the left hand side is expressible as product of two linear factors, which is possible if abc + 2fgh – af2 – bg2 – ch2 = 0

or ...(2)

The expression (2) is also called the DISCRIMINANT of the equation (1).

IMPORTANT RESULTS

- Angle Î¸ between the lines is given by
- The point of intersection of the pair of lines is.
- The equation of the pair of angular bisectors is given by , where (x1, y1) is the point of intersection of two lines. [It is obtained by replacing x by x – x1 and y by y – y1 in the equation

### AN ALTERNATE METHOD FOR FINDING POINT OF INTERSECTION

Let ax2 + 2hxy + by2 + 2gx + 2fy + c or represents a pair of straight lines.

Find, i.e. differentiate with respect to x treating y as a constant.

Then

(ax2 + 2hxy + by2 + 2gx + 2fy + c = 0) = 2ax + 2hy + 2g

Similarly find (Ï† is differentiated with respect to y treating x as a constant)

Solve the equations to get the required point of intersection.

Thus solving the equations ax + hy + g = 0 and hx + by + f = 0, we get

and, which gives the abscissa and ordinate of the point of intersection.

### SEPARATION OF THE EQUATIONS OF LINES

Let the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents the straight lines y = m1x + c1 and y = m2x + c2.

Then, ax2 + 2hxy + by2 + 2gx + 2fy + c

=b (y – m1x – c1) (y – m2x – c2)

Comparing the coefficients, we get

USEFUL RESULTS

- The two lines are parallel if h2 – ab = 0
- The two lines are perpendicular if a + b = 0
- The two lines are coincident if h2 – ab = g2 – ac = f2 – bc = 0
- Distance between the parallel lines : If the two lines represented by eqn. (1) are parallel, then the distance between the two parallel lines is given by

- The two pairs of straight lines ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 and ax2 + by2 + 2hxy = 0 form a
- square if (a – b) fg + h (f2 – g2) = 0, a + b = 0
- rectangle if (a – b) fg + h (f2 – g2) ≠ 0, a + b = 0
- rhombus if (a – b) fg + h (f2 – g2) = 0, a + b ≠ 0
- parallelogram if (a – b) fg + h (f2 – g2) ≠ 0, a + b ≠ 0

### EQUATION OF THE PAIR OF STRAIGHT LINES JOINING THE ORIGIN TO THE POINTS OF INTERSECTION OF A CURVE AND A STRAIGHT LINE

Let the equation of the given curve be

and the equation of the straight line be

L : lx + my + n = 0.

To find the equation of the pair of straight lines joining the points of intersection A and B of the curve S = 0 and the line L = 0 with the origin O, we homogenise the equation S = 0, with the help of L = 0. For this, we write

and

⇒ ax2 + 2hxy + by2 + 2 (gx + fy)

which is homogeneous equation of the second degree and gives the required pair of straight lines.

PROCEDURE

- Divide the equation of the line L = 0 by the negative of constant term and write it as
- Multiply the single degree terms by 1 and the absolute terms by 12 in the equation of the curve and put the value of 1 as obtained in above step 1.

### TRANSLATION OF AXES

A change of origin without changing the direction of coordinate axes is called a translation of axes.

Let xoy represent the initial system. The origin is shifted to a new point O maintaining the axes parallel to original position then the new system is XOY.

Let the coordinates of O with respect to initial system are (h, k).

Suppose P is a point in the plane whose coordinates are (x, y) initially and (X, Y) in the translated system.

Then x = X + h and y = Y + k

Therefore the equation of any curve f(x, y) = 0 will become f(X + h, Y + k), where reference is taken on the translated axes.

### ROTATION OF AXES

Let xoy represent the initial system. Now the axes are rotated by Î¸ angle anticlockwise keeping the origin unchanged, then the new system is XOY.

Let P is a point in the plane whose coordinates are (x, y) initially and (X, Y) in the new system.

Then, x = Xcos Î¸ – Y sin Î¸; y = X sin Î¸ + Y cos Î¸

and X = x cos Î¸ + y sin Î¸; Y = –x sin Î¸ + y cos Î¸

The above result can be remembered by the following matrix equation.

and