General Equation of Family of Lines , Concurrency of Straight Lines , Solved Example

The general equation of the family of lines through the point of intersection of two given lines is L + λL’ = 0 , where L = 0 and L’ = 0 are the two given lines, and λ is a parameter.
Conversely, any line of the form L1 + λ L2 = 0 passes through a fixed point which is the point of intersection of the lines L1 = 0 and L2 = 0.

∎ The family of lines perpendicular to a given line ax + by + c = 0 is given by bx − ay + k = 0, where k is a parameter.

∎ The family of lines parallel to a given line ax + by + c = 0 is given by ax + by + k = 0, where k is a parameter.

Illustration : Show that all the chords of the curve 3x2 − y2 − 2x + 4y = 0 , which subtend a right angle at the origin pass through a fixed point. Find that point.

Solution: Let the equation of chord be lx + my = 1.

So equation of pair of straight line joining origin to the points of intersection of chord and curve.

3x2 − y2 − 2x(lx + my) + 4y(lx + my) = 0, which subtends right angle at origin.

=> (3 − 2l + 4m − 1) = 0 => l = 2m + 1

Hence chord becomes (2m + 1)x + my = 1

x − 1 + m(2x + y) = 0
L1          L2

Which will pass through point of intersection of L1 = 0 and L2 = 0

=> x = 1 , y = −2. Hence fixed point is (1, −2).

One Parameter Family of Straight Lines :

If a linear expression L1 contains an unknown coefficient, then the line L1 = 0 can not be a fixed line. Rather it represents a family of straight lines known as one parameter family of straight lines. e.g. family of lines parallel to the x-axis i.e. y = c and family of straight lines passing through the origin i.e. y = mx.

Each member of the family passes through a fixed point. We have two methods to find the fixed point.

Method (i):

Let the family of straight lines be of the form ax + by + c = 0 where a, b, c are variable parameters satisfying the condition a l + b m + c n = 0 , where l , m , n are given and n ≠ 0.
Rewriting the condition as $\large a(\frac{l}{n}) + b(\frac{m}{n}) + c = 0$ , and comparing with the given family of straight lines, we find that each member of it passes through the fixed point $\large (\frac{l}{n} , \frac{m}{n})$

Illustration : If the algebraic sum of perpendiculars from n given points on a variable straight line is zero then prove that the variable straight line passes through a fixed point.

Solution: Let n given points be (xi , yi) where i = 1, 2 …. n and the variable straight line is ax + by + c = 0. Given that
$\large \Sigma_{i=1}^{n} \frac{a x_i + b y_i + c}{\sqrt{a^2 + b^2}} = 0$

⇒ a Σxi + b Σyi + c n = 0

$\large a \frac{\Sigma x_i}{n} + b \frac{\Sigma y_i}{n} + c = 0$

Hence the variable straight line always passes through the fixed point

$\large (\frac{\Sigma x_i}{n} , \frac{\Sigma y_i}{n} )$

Method (ii):

If a family of straight lines can be written as L1 + λL2 = 0 where L1 , L2 are two fixed lines and λ is a parameter, then each member of it will pass through a fixed point given by point of intersection of L1 = 0 and L2 = 0.

Note:
∎ If L1 = 0 and L2 = 0 are parallel lines , they will meet at infinity.

Illustration : Prove that each member of the family of straight lines
(3sinθ + 4cosθ)x + (2sinθ − 7cosθ)y + (sinθ + 2cosθ) = 0 ( θ is a parameter )
passes through a fixed point.

Solution: The given family of straight lines can be rewritten as

(3x + 2y + 1) sinθ + (4x − 7y + 2) cosθ = 0

or, (4x − 7y + 2) + tanθ (3x + 2y + 1) = 0 which is of the form L1 + λ L2 = 0

Hence each member of it will pass through a fixed point which is the intersection of 4x − 7y + 2 = 0 and 3x + 2y +1 = 0 i.e. (−11/29 , 2/29)

Concurrency of Straight Lines

The condition for 3 lines a1x + b1y + c1 = 0 , a2x + b2y + c2 = 0 , a3x + b3y + c3 = 0 to be concurrent is

(i) $\large \left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right| = 0$

(ii) There exist 3 constants l , m , n (not all zero at the same time) such that
L1 + mL2 + nL3 = 0 , where L1 = 0 , L2 = 0 and L3 = 0 are the three given straight lines.

(iii) The three lines are concurrent if any one of the lines passes through the point of intersection of the other two lines.

Exercise :
i) Find the equation of a straight line passing through the point (4, 5) and equally inclined to the lines 3x = 4y + 7 and 5y = 12x + 6.

ii) Find the equation of the bisector of the obtuse angle between the lines x − 2y + 4 = 0 and 4x − 3y + 2 = 0.

iii) Find the equations of the lines which pass through the point of intersection of the lines 4x − 3y − 1 = 0 and 2x − 5y + 3 = 0 and are equally inclined to the axes.

iv) A straight line cuts intercepts from the axes of co-ordinates the sum of the reciprocals of which is constant. Show that it always passes through a fixed point.