Family of Lines and Position of Points

Introduction

This section deals with advanced concepts used to handle problems involving multiple lines and the relative position of points, which are very common in competitive exams.

1. Family of Lines

A family of lines is an infinite set of lines that share a common geometric property. The most important case is the family of lines passing through the point of intersection of two given lines.

• Equation of the Family: Given two lines $L_1: A_1x + B_1y + C_1 = 0$ and $L_2: A_2x + B_2y + C_2 = 0$, the equation of any line passing through their point of intersection is given by:

$$L_1 + \lambda L_2 = 0$$

where $\lambda$ is a real number called a parameter. A specific condition (like another point the line passes through, or a required slope) is used to find the unique value of $\lambda$ for the desired line.

2. Position of a Point Relative to a Line

We can determine if two points lie on the same side or on opposite sides of a given line without needing to graph them.

• The Rule: For a line $L: Ax + By + C = 0$ and two points $P(x_1, y_1)$ and $Q(x_2, y_2)$:

  1. Substitute the coordinates of the first point into the line's expression: $L(P) = Ax_1 + By_1 + C$.

  2. Substitute the coordinates of the second point: $L(Q) = Ax_2 + By_2 + C$.

  3. Compare the signs of the results:

     • If $L(P)$ and $L(Q)$ have the same sign, the points lie on the same side of the line.

     • If $L(P)$ and $L(Q)$ have opposite signs, the points lie on opposite sides of the line.

Example 1: Using the Family of Lines

Question: Find the equation of the line passing through the intersection of the lines $2x+y-1=0$ and $x-3y+1=0$ and also passing through the point (1,1).

Solution:

Instead of finding the intersection point first, we can use the family of lines method.

Step 1: Write the equation for the family of lines.

$L_1 + \lambda L_2 = 0 \implies (2x+y-1) + \lambda(x-3y+1) = 0$.

Step 2: Use the given point to find the value of $\lambda$.

Since the desired line passes through (1,1), we substitute $x=1, y=1$ into the equation:

$(2(1)+1-1) + \lambda(1-3(1)+1) = 0$

$2 + \lambda(-1) = 0 \implies \mathbf{\lambda=2}$.

Step 3: Substitute $\lambda=2$ back into the family equation.

$(2x+y-1) + 2(x-3y+1) = 0$

$2x+y-1 + 2x-6y+2 = 0$

$\mathbf{4x - 5y + 1 = 0}$.

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Example 2: Family of Lines with a Slope Condition

Question: Find the equation of the line passing through the intersection of $x-y+1=0$ and $2x+y-4=0$ and perpendicular to the line $x+3y=1$.

Solution:

Step 1: Write the equation for the family of lines and find its slope in terms of $\lambda$.

$(x-y+1) + \lambda(2x+y-4) = 0$

Rearranging: $(1+2\lambda)x + (-1+\lambda)y + (1-4\lambda) = 0$.

The slope of this line is $m_1 = -\frac{A}{B} = -\frac{1+2\lambda}{-1+\lambda} = \frac{1+2\lambda}{1-\lambda}$.

Step 2: Find the slope of the second line and apply the perpendicularity condition.

The slope of $x+3y=1$ is $m_2 = -1/3$. For the lines to be perpendicular, $m_1 m_2 = -1$.

$$(\frac{1+2\lambda}{1-\lambda}) (-\frac{1}{3}) = -1 \implies 1+2\lambda = 3(1-\lambda) = 3-3\lambda$$

$$5\lambda = 2 \implies \mathbf{\lambda=2/5}$$

Step 3: Substitute $\lambda=2/5$ back into the family equation.

$(x-y+1) + \frac{2}{5}(2x+y-4) = 0$

Multiply by 5: $5(x-y+1) + 2(2x+y-4) = 0$

$5x-5y+5 + 4x+2y-8=0 \implies 9x-3y-3=0$, which simplifies to $3x-y-1=0$.

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Example 3: Position of a Point

Question: Determine if the points (2,1) and (-3,5) lie on the same side or opposite sides of the line $2x-3y+6=0$.

Solution:

Let the line be $L(x,y) = 2x-3y+6$.

Step 1: Evaluate L at the first point, P(2,1).

$L(2,1) = 2(2) - 3(1) + 6 = 4-3+6 = \mathbf{7}$ (Positive).

Step 2: Evaluate L at the second point, Q(-3,5).

$L(-3,5) = 2(-3) - 3(5) + 6 = -6-15+6 = \mathbf{-15}$ (Negative).

Since the signs of the results are opposite, the points lie on opposite sides of the line.