Q.A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.
Imagine three friends standing in a field. Each friend holds a long, straight rope stretched tight. If all three ropes pass through exactly the same point — say, a flagpole in the centre — then the ropes are concurrent. That common point is called the point of concurrency.
In geometry, when three or more lines all pass through a single point, we say they are concurrent lines. That point is their concurrency point.
Note
Two lines are always concurrent (unless they are parallel) — they meet at exactly one point. The interesting case is three or more lines. Do they all happen to pass through the same spot?
The Intuition Behind the Condition
Suppose you have three lines:
L1:a1x+b1y+c1=0
L2:a2x+b2y+c2=0
L3:a3x+b3y+c3=0
If they are concurrent, there exists some point (x0,y0) that satisfies all three equations at once. That means (x0,y0) is a common solution.
Now, think about it this way:
The first two lines L1 and L2 intersect at some point P (unless they are parallel). For the three lines to be concurrent, L3 must also pass through that same point P. So the condition boils down to: the point of intersection of any two lines must lie on the third line.
That is the simplest way to check concurrency: solve two equations, get the intersection, and plug it into the third equation. If it satisfies, the lines are concurrent.
The Precise Algebraic Condition
There is a cleaner, more powerful condition using determinants — it avoids solving for the intersection explicitly.
Three lines a1x+b1y+c1=0, a2x+b2y+c2=0, a3x+b3y+c3=0 are concurrent if and only if
a1a2a3b1b2b3c1c2c3=0
This determinant being zero is the necessary and sufficient condition for concurrency of three lines.
Watch out
This condition assumes that no two of the lines are parallel. If L1 and L2 are parallel, they never meet, so the three lines cannot be concurrent (unless all three are the same line, which is a degenerate case). The determinant condition will still give zero in that parallel case, but the lines are not concurrent — they are parallel. So always check that the lines actually intersect pairwise first.
Why Does the Determinant Work?
Here is the reasoning in plain steps:
For concurrency, there must exist (x0,y0) such that:
a1x0+b1y0+c1=0
a2x0+b2y0+c2=0
a3x0+b3y0+c3=0
Think of these as three equations in three unknowns: x0, y0, and the constant 1. Yes, the constant 1 is treated as a variable here.
For a non-trivial solution to exist (i.e., a solution where the "variables" are not all zero), the determinant of the coefficient matrix must be zero. That is a standard result from linear algebra: a homogeneous system has a non-zero solution only when the determinant is zero.
The determinant being zero is exactly the condition that the three equations are linearly dependent — meaning one equation can be written as a combination of the other two. That is another way to say: the third line passes through the intersection of the first two.
Tip
For quick checks in exams, use the determinant. But if the numbers are simple, solving two equations and substituting into the third is often faster and less error-prone.
When the sum of reciprocals of intercepts is constant, say a1+b1=k, the line always passes through the fixed point (k1,k1).
The problem asks us to prove that a family of lines satisfying a particular constraint all share a common point. This is a beautiful example of how algebraic constraints translate into geometric properties.
Understanding the constraint
A line with intercepts a and b on the x- and y-axes respectively has the intercept form equation:
ax+by=1
The condition states that a1+b1 remains constant as the line moves. Let's call this constant k:
a1+b1=k
where k is some fixed non-zero real number.
The key insight is this: if we can find coordinates (h,k) that satisfy the line equation for every choice of a and b obeying our constraint, then (h,k) is a fixed point through which all such lines pass.
Finding the fixed point
1. Express the constraint in terms of intercepts
From the given condition:
a1+b1=k
Rearranging:
aba+b=k⟹a+b=kab
2. Rewrite the line equation using the constraint
The line equation is ax+by=1. Multiply both sides by ab:
bx+ay=ab
From our constraint, ab=ka+b. Substituting:
bx+ay=ka+b
Multiply through by k:
k(bx+ay)=a+b
Rearranging:
kbx+kay=a+b
kbx−b=a−kay
b(kx−1)=a(1−ky)
3. Identify the fixed point
For this equation to hold for all valid pairs (a,b) satisfying our constraint, we need both sides to vanish independently. This happens when: