Welcome to this detailed exploration of the "Intersection of Lines" โ a fundamental concept in Coordinate Geometry that forms the bedrock for numerous advanced problems in JEE. Understanding how lines interact in a plane is crucial, and we'll dive deep into all possible scenarios, methods to find intersection points, and advanced applications like families of lines and concurrency.
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1. Introduction to Intersection of Lines
In a two-dimensional Cartesian plane, a straight line represents an infinite set of points satisfying a linear equation. When we talk about the "intersection of lines," we are essentially looking for points that lie on *two or more* lines simultaneously. Geometrically, this point (or points) is where the lines cross or coincide.
Consider two distinct lines, L1 and L2. When drawn on a plane, there are only three possible relationships they can have:
1.
Intersecting Lines: They cross each other at exactly
one unique point. This is the most common scenario we deal with.
2.
Parallel and Distinct Lines: They never meet, no matter how far they are extended. In this case, there is
no point of intersection.
3.
Coincident Lines: They are essentially the same line, lying directly on top of each other. This means they share
infinitely many points of intersection.
Our primary goal is to understand these conditions and, for the first case, how to find that unique point of intersection.
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2. Finding the Point of Intersection of Two Lines (Unique Case)
When two lines intersect at a unique point, that point must satisfy the equations of both lines simultaneously. This means finding the point of intersection is equivalent to solving a system of two linear equations in two variables (x and y).
Let the equations of two lines be:
Line 1 (L1): $A_1x + B_1y + C_1 = 0$
Line 2 (L2): $A_2x + B_2y + C_2 = 0$
There are several standard methods to solve such a system:
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Method 1: Substitution Method
1. Express one variable (say, y) in terms of the other (x) from one equation.
2. Substitute this expression into the second equation.
3. Solve the resulting single-variable equation for x.
4. Substitute the value of x back into the expression from step 1 to find y.
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Method 2: Elimination Method
1. Multiply one or both equations by suitable constants such that the coefficients of one variable (say, y) become equal in magnitude but opposite in sign.
2. Add the two equations to eliminate that variable.
3. Solve the resulting single-variable equation for the remaining variable (x).
4. Substitute the value of x back into either original equation to find y.
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Method 3: Cross-Multiplication Method (General Formula Derivation)
This method is particularly efficient for deriving a general formula for the intersection point.
Given:
$A_1x + B_1y + C_1 = 0$
$A_2x + B_2y + C_2 = 0$
Using the cross-multiplication rule:
$frac{x}{B_1C_2 - B_2C_1} = frac{y}{C_1A_2 - C_2A_1} = frac{1}{A_1B_2 - A_2B_1}$
Provided that $A_1B_2 - A_2B_1
eq 0$ (which implies a unique solution, i.e., intersecting lines), the coordinates of the intersection point $(x, y)$ are:
The point of intersection $(x,y)$ is given by:
$x = frac{B_1C_2 - B_2C_1}{A_1B_2 - A_2B_1}$
$y = frac{C_1A_2 - C_2A_1}{A_1B_2 - A_2B_1}$
This formula is very powerful as it directly gives the intersection point for any two lines in general form. Remember this formula, as it can save time in JEE problems.
Example 1: Finding the point of intersection
Find the point of intersection of the lines $2x - 3y + 5 = 0$ and $3x + y - 2 = 0$.
Solution:
We can use the elimination method:
1. Equation 1: $2x - 3y + 5 = 0$
2. Equation 2: $3x + y - 2 = 0$
Multiply Equation 2 by 3 to make the y coefficients suitable for elimination:
$3 imes (3x + y - 2 = 0) implies 9x + 3y - 6 = 0$ (Equation 3)
Now, add Equation 1 and Equation 3:
$(2x - 3y + 5) + (9x + 3y - 6) = 0$
$11x - 1 = 0$
$11x = 1 implies x = frac{1}{11}$
Substitute $x = frac{1}{11}$ into Equation 2:
$3left(frac{1}{11}
ight) + y - 2 = 0$
$frac{3}{11} + y - 2 = 0$
$y = 2 - frac{3}{11} = frac{22 - 3}{11} = frac{19}{11}$
Therefore, the point of intersection is $left(frac{1}{11}, frac{19}{11}
ight)$.
JEE Tip: For objective questions, if the coefficients are straightforward, the general formula (cross-multiplication method) can be faster. Practice all methods to choose the most efficient one for a given problem.
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3. Conditions for Intersection, Parallelism, and Coincidence
Let the two lines be $L_1: A_1x + B_1y + C_1 = 0$ and $L_2: A_2x + B_2y + C_2 = 0$.
The nature of their intersection depends on the ratios of their coefficients.
1.
Intersecting Lines (Unique Solution):
The lines intersect at a unique point if and only if their slopes are different.
Slope of $L_1 = m_1 = -frac{A_1}{B_1}$
Slope of $L_2 = m_2 = -frac{A_2}{B_2}$
For unique intersection, $m_1
eq m_2 implies -frac{A_1}{B_1}
eq -frac{A_2}{B_2} implies frac{A_1}{B_1}
eq frac{A_2}{B_2}$.
This can be rewritten as
$A_1B_2 - A_2B_1
eq 0$.
This condition also means that the denominator in the cross-multiplication formula is non-zero, allowing for a unique solution.
2.
Parallel and Distinct Lines (No Solution):
The lines are parallel if their slopes are equal, but they have different y-intercepts.
$m_1 = m_2 implies frac{A_1}{B_1} = frac{A_2}{B_2}$ (or $A_1B_2 - A_2B_1 = 0$)
And their y-intercepts are different.
This leads to the condition:
$frac{A_1}{A_2} = frac{B_1}{B_2}
eq frac{C_1}{C_2}$.
In this case, the system of equations has no solution. Geometrically, the lines never meet.
3.
Coincident Lines (Infinite Solutions):
The lines are coincident if they are essentially the same line. This means their slopes are equal, and their y-intercepts are also equal.
$frac{A_1}{A_2} = frac{B_1}{B_2} = frac{C_1}{C_2}$.
In this scenario, every point on one line is also a point on the other, leading to infinitely many solutions.
Example 2: Determining the nature of intersection
Determine whether the following pairs of lines are intersecting, parallel, or coincident:
a) $4x + 6y - 7 = 0$ and $2x + 3y + 1 = 0$
b) $x - 2y + 3 = 0$ and $3x - 6y + 9 = 0$
c) $x + y - 1 = 0$ and $2x - y + 4 = 0$
Solution:
a) $L_1: 4x + 6y - 7 = 0 implies A_1=4, B_1=6, C_1=-7$
$L_2: 2x + 3y + 1 = 0 implies A_2=2, B_2=3, C_2=1$
Check ratios:
$frac{A_1}{A_2} = frac{4}{2} = 2$
$frac{B_1}{B_2} = frac{6}{3} = 2$
$frac{C_1}{C_2} = frac{-7}{1} = -7$
Since $frac{A_1}{A_2} = frac{B_1}{B_2}
eq frac{C_1}{C_2}$, the lines are
parallel and distinct.
b) $L_1: x - 2y + 3 = 0 implies A_1=1, B_1=-2, C_1=3$
$L_2: 3x - 6y + 9 = 0 implies A_2=3, B_2=-6, C_2=9$
Check ratios:
$frac{A_1}{A_2} = frac{1}{3}$
$frac{B_1}{B_2} = frac{-2}{-6} = frac{1}{3}$
$frac{C_1}{C_2} = frac{3}{9} = frac{1}{3}$
Since $frac{A_1}{A_2} = frac{B_1}{B_2} = frac{C_1}{C_2}$, the lines are
coincident.
c) $L_1: x + y - 1 = 0 implies A_1=1, B_1=1, C_1=-1$
$L_2: 2x - y + 4 = 0 implies A_2=2, B_2=-1, C_2=4$
Check ratios:
$frac{A_1}{A_2} = frac{1}{2}$
$frac{B_1}{B_2} = frac{1}{-1} = -1$
Since $frac{A_1}{A_2}
eq frac{B_1}{B_2}$ (i.e., $frac{1}{2}
eq -1$), the lines are
intersecting.
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4. Lines Passing Through the Intersection of Two Given Lines (Family of Lines)
This is a critically important concept for JEE Advanced problems. Often, you'll need to find the equation of a line that passes through the intersection of two given lines and also satisfies some other condition. Instead of finding the intersection point first and then using the point-slope form, there's a more elegant and efficient approach.
Consider two lines $L_1 = 0$ and $L_2 = 0$:
$L_1: A_1x + B_1y + C_1 = 0$
$L_2: A_2x + B_2y + C_2 = 0$
The equation of any straight line passing through the point of intersection of $L_1 = 0$ and $L_2 = 0$ is given by:
$L_1 + lambda L_2 = 0$
or
$(A_1x + B_1y + C_1) + lambda (A_2x + B_2y + C_2) = 0$
where $lambda$ (lambda) is any real number.
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Why does this work?
1.
It's a Line: Expanding the equation $(A_1x + B_1y + C_1) + lambda (A_2x + B_2y + C_2) = 0$, we get $(A_1 + lambda A_2)x + (B_1 + lambda B_2)y + (C_1 + lambda C_2) = 0$. This is clearly a linear equation in x and y (of the form $Ax + By + C = 0$), so it represents a straight line.
2.
It Passes Through the Intersection: Let $(x_0, y_0)$ be the point of intersection of $L_1=0$ and $L_2=0$. This means $(x_0, y_0)$ satisfies both equations:
$A_1x_0 + B_1y_0 + C_1 = 0$
$A_2x_0 + B_2y_0 + C_2 = 0$
Now, substitute $(x_0, y_0)$ into the family equation:
$(A_1x_0 + B_1y_0 + C_1) + lambda (A_2x_0 + B_2y_0 + C_2) = 0$
$0 + lambda (0) = 0$
$0 = 0$
Since this equation holds true for any value of $lambda$, any line represented by $L_1 + lambda L_2 = 0$ must pass through the point of intersection $(x_0, y_0)$.
Important Note: This family of lines includes every line passing through the intersection point EXCEPT for the line $L_2=0$ itself (because $L_2=0$ would require $lambda$ to be infinite, or effectively, for $L_1$ to be zero and then a constant times $L_2$ be zero. If you write it as $k_1 L_1 + k_2 L_2 = 0$, then both $L_1=0$ (when $k_2=0$) and $L_2=0$ (when $k_1=0$) are included). In most practical problems, $L_2=0$ is implicitly handled, or it's obvious from the context which specific line is being sought.
To find a specific line from this family, you will always be given one additional condition, which allows you to determine the value of $lambda$.
Example 3: Line through intersection and another point
Find the equation of the line passing through the intersection of lines $x + y - 5 = 0$ and $2x - y + 4 = 0$, and also passing through the point $(1, 2)$.
Solution:
Let $L_1: x + y - 5 = 0$ and $L_2: 2x - y + 4 = 0$.
The equation of a line passing through their intersection is $L_1 + lambda L_2 = 0$:
$(x + y - 5) + lambda (2x - y + 4) = 0$
This line also passes through the point $(1, 2)$. Substitute $x=1$ and $y=2$ into the equation:
$(1 + 2 - 5) + lambda (2(1) - 2 + 4) = 0$
$(-2) + lambda (2 - 2 + 4) = 0$
$-2 + lambda (4) = 0$
$4lambda = 2 implies lambda = frac{1}{2}$
Now substitute $lambda = frac{1}{2}$ back into the family equation:
$(x + y - 5) + frac{1}{2} (2x - y + 4) = 0$
Multiply by 2 to clear the fraction:
$2(x + y - 5) + (2x - y + 4) = 0$
$2x + 2y - 10 + 2x - y + 4 = 0$
$4x + y - 6 = 0$
This is the required equation of the line.
Example 4: Line through intersection and parallel to another line
Find the equation of the line passing through the intersection of lines $3x - y = 0$ and $x + y - 2 = 0$, and parallel to the line $x + 2y = 0$.
Solution:
Let $L_1: 3x - y = 0$ and $L_2: x + y - 2 = 0$.
The equation of a line passing through their intersection is $L_1 + lambda L_2 = 0$:
$(3x - y) + lambda (x + y - 2) = 0$
Rearrange this into the general form $Ax + By + C = 0$:
$(3 + lambda)x + (-1 + lambda)y - 2lambda = 0$
The slope of this line is $m = -frac{ ext{Coefficient of x}}{ ext{Coefficient of y}} = -frac{(3 + lambda)}{(-1 + lambda)} = frac{3 + lambda}{1 - lambda}$.
The required line is parallel to $x + 2y = 0$.
The slope of $x + 2y = 0$ is $m_{parallel} = -frac{1}{2}$.
Since the lines are parallel, their slopes must be equal:
$frac{3 + lambda}{1 - lambda} = -frac{1}{2}$
$2(3 + lambda) = -1(1 - lambda)$
$6 + 2lambda = -1 + lambda$
$2lambda - lambda = -1 - 6$
$lambda = -7$
Now substitute $lambda = -7$ back into the family equation:
$(3x - y) - 7(x + y - 2) = 0$
$3x - y - 7x - 7y + 14 = 0$
$-4x - 8y + 14 = 0$
Dividing by -2:
$2x + 4y - 7 = 0$
This is the required equation of the line.
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5. Concurrency of Three Lines
Three or more lines are said to be
concurrent if they all pass through a single common point.
To check if three given lines are concurrent:
Let the three lines be:
$L_1: A_1x + B_1y + C_1 = 0$
$L_2: A_2x + B_2y + C_2 = 0$
$L_3: A_3x + B_3y + C_3 = 0$
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Method 1: Direct Verification
1. Find the point of intersection of any two lines (say, $L_1$ and $L_2$) using methods discussed earlier.
2. Substitute the coordinates of this intersection point into the equation of the third line ($L_3$).
3. If the point satisfies $L_3=0$, then the three lines are concurrent. Otherwise, they are not.
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Method 2: Using Determinants (JEE Advanced Focus)
Three lines $A_1x + B_1y + C_1 = 0$, $A_2x + B_2y + C_2 = 0$, and $A_3x + B_3y + C_3 = 0$ are concurrent if and only if the determinant of their coefficients is zero.
$egin{vmatrix} A_1 & B_1 & C_1 \ A_2 & B_2 & C_2 \ A_3 & B_3 & C_3 end{vmatrix} = 0$
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Derivation of the Determinant Condition:
If the three lines are concurrent, there exists a unique point $(x_0, y_0)$ that satisfies all three equations:
$A_1x_0 + B_1y_0 + C_1 = 0$
$A_2x_0 + B_2y_0 + C_2 = 0$
$A_3x_0 + B_3y_0 + C_3 = 0$
This is a system of three linear equations in $x_0$ and $y_0$. For a non-trivial solution (i.e., not all coefficients being zero in a specific way) for $x_0$ and $y_0$ to exist, and for consistency, the determinant of the coefficients (when viewed as a system for $x_0, y_0, 1$) must be zero.
Alternatively, consider that if they are concurrent, the third line must be part of the family of lines passing through the intersection of the first two.
So, $L_3$ can be expressed as $L_1 + lambda L_2 = 0$ for some $lambda$.
This means $(A_3x + B_3y + C_3)$ is linearly dependent on $(A_1x + B_1y + C_1)$ and $(A_2x + B_2y + C_2)$.
The condition for linear dependence of three linear forms $L_1, L_2, L_3$ is that the determinant of their coefficients is zero. This is a direct application from linear algebra.
Example 5: Checking for concurrency
Check if the lines $x - y + 1 = 0$, $2x + y - 10 = 0$, and $3x - 2y + 4 = 0$ are concurrent.
Solution:
Method 1: Direct Verification
1. Find the intersection of $L_1: x - y + 1 = 0$ and $L_2: 2x + y - 10 = 0$.
Adding the two equations:
$(x - y + 1) + (2x + y - 10) = 0$
$3x - 9 = 0 implies 3x = 9 implies x = 3$
Substitute $x=3$ into $L_1$:
$3 - y + 1 = 0 implies 4 - y = 0 implies y = 4$
The intersection point of $L_1$ and $L_2$ is $(3, 4)$.
2. Substitute $(3, 4)$ into $L_3: 3x - 2y + 4 = 0$.
$3(3) - 2(4) + 4$
$9 - 8 + 4 = 5$
Since $5
eq 0$, the point $(3, 4)$ does *not* satisfy the equation of $L_3$.
Therefore, the three lines are
not concurrent.
Method 2: Using Determinants
Coefficients are:
$L_1: (1, -1, 1)$
$L_2: (2, 1, -10)$
$L_3: (3, -2, 4)$
Calculate the determinant:
$egin{vmatrix} 1 & -1 & 1 \ 2 & 1 & -10 \ 3 & -2 & 4 end{vmatrix}$
$= 1(1 imes 4 - (-10) imes (-2)) - (-1)(2 imes 4 - (-10) imes 3) + 1(2 imes (-2) - 1 imes 3)$
$= 1(4 - 20) + 1(8 + 30) + 1(-4 - 3)$
$= 1(-16) + 1(38) + 1(-7)$
$= -16 + 38 - 7$
$= 22 - 7 = 15$
Since the determinant is $15
eq 0$, the lines are
not concurrent. This confirms the result from Method 1.
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6. JEE Focus: Advanced Applications
The concepts of intersection and family of lines are foundational. In JEE problems, they are rarely asked in isolation. Instead, they are combined with other concepts:
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Distance Formulas: Find a line passing through an intersection point such that its distance from a given point is maximized/minimized.
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Angles between Lines: Find a line through an intersection that makes a specific angle with another line.
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Reflection: Problems involving light rays reflecting off a line, where the incident and reflected rays meet at a point on the line.
*
Area of Triangles: Calculating the area of a triangle formed by three lines (by finding their three intersection points).
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Locus Problems: Finding the locus of a point related to intersecting lines (e.g., orthocenter, circumcenter, incenter, excenters, when they are defined by a parameter).
*
Conic Sections: Tangents to conic sections often involve lines passing through specific points, which might be an intersection of other lines.
The "family of lines" concept ($L_1 + lambda L_2 = 0$) is particularly powerful for JEE problems because it allows you to represent the desired line without explicitly finding the intersection point, thereby simplifying calculations. Master its application!
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