(a1x +b1y +c1) (a2x +b2y+ c2) = 0

ax² +2hxy+by²+2gx+2fy+c = 0 is joint equation of a pair of straight lines.

The equation ax² +2hxy+by²+2gx+2fy+c = 0 is known as general equation of second degree.

The equation ax² +2hxy+by² = 0 is known as homogeneous equation of second degree.

In a homogeneous equation of second degree, the sum of indices (exponents) of x and y in each term is equal to 2.

The homogeneous equation of second degree ax² +2hxy+by² = 0 represents a joint equation of two straight lines passing through the origin if h²≥ab.

If y = m1x and y = m2x are the lines represented by a homogeneous equation of second degree ax² +2hxy+by² = 0, then

(i) m1 =m2 = -2h/b

(ii) m1m2 = a/b

The angle θ between the pair of lines represented the homogeneous equation of second degree ax² +2hxy+by² = 0 is given by

tan θ = [2√(h² –ab)]/(a+b)

If θ = 0, which means h² = ab lines are coincident.

Lines are perpendicular means θ = π/2, tan θ = ∞, and cot θ = 0.

This means a+b = 0 or a = -b

Coefficient of x² = coefficient of y²

ax² +2hxy+by²+2gx+2fy+c = 0 will represent a pair of straight lines if the determinant

|a h g|

|h b f|

|g f c|

= 0

Expanding the determinant

abc +2fgh -af² -bg² -ch² = 0

Angle θ between the lines represented by the general second degree equation ax² +2hxy+by²+2gx+2fy+c = 0 is given by

tan θ = [2 √(h² – ab)]/(a+b)

**Algorithm to find separate equations of lines**in ax² +2hxy+by²+2gx+2fy+c = 0

Step 1. Find factors for the homogeneous part ax² +2hxy+by². Let the factors be

(a1x +b1y ) and (a2x +b2y )

Step 2.Add constants c1 and c2 to them. (a1x +b1y +c1) and (a2x +b2y +c2).

Step 3. Multiply (a1x +b1y +c1) and (a2x +b2y +c2) and compare with ax² +2hxy+by²+2gx+2fy+c to obtain equations in c1 and c2.

Step 4. Solve the equations and get values of c1 and c2.

**Bisectors of the angle between the lines given by a homogeneous equation**

The jont equation of the bisectors of the angles between the lines represented by ax² +2hxy+by² = 0 is given by

(x²-y²)/(a-b) = xy/h

**Equations for bisectors**of the lines represented by ax² +2hxy+by²+2gx+2fy+c = 0

[(x-x1) ² – (y-y1) ²]/(a-b) = (x-x1)(y-y1)/h

where x1,y1 is the point of intersection of the lines represented by the given equation.

**Algorithm to find the joint equation of lines joining the origin to the points of intersection of a line and a curve.**

Step 1. Take all terms of x and y in the equation of the line on LHS and the constant term on RHS, then divide both sides by the this constant on RHS, so that RHS becomes unity.

Step 2. Multiply the first degree terms in the equation of the curve by the LHS obtained in step 1 and the constant term by the square of the LHS obtained in the step 1, keeping the second degree terms unchanged. The required equation is obtained.

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