JEE Advanced 1978

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Animated Solution for Mathematics - Straight Lines: A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

Visualized Solution

Visualizing the Sliding Rod

Let the two perpendicular lines be the coordinate axes, $O X$ and $O Y$ .
A line segment $A B$ of constant length $l$ slides such that its ends $A$ and $B$ always remain on the $x$ -axis and $y$ -axis respectively.
Imagine a ladder sliding down a wall—this is the exact physical scenario!

Coordinates of the Moving Ends

Let the coordinates of the end points be $A (a, 0)$ and $B (0, b)$ .
Here, $a$ and $b$ represent the variable intercepts on the axes.
As the rod slides, the values of $a$ and $b$ continuously change, but they are constrained by the length of the rod.

Applying the Length Constraint

The length of the segment $A B$ is constant and equal to $l$ .
Using the distance formula or Pythagoras theorem in right-angled $△ O A B$ :
$a^{2} + b^{2} = l^{2}$

Locating the Point $P (x, y)$

Let $P (x, y)$ be the point that divides the segment $A B$ in the ratio $1 : 2$ .
This means $A P : P B = 1 : 2$ .
We need to find the relation between $x$ and $y$ that remains true for all positions of the rod.

Section Formula for $x$

By the section formula, the $x$ -coordinate of $P$ is:
$x = \frac{m \cdot x _{2} + n \cdot x _{1}}{m + n}$
Substituting $m = 1, n = 2$ , $A (a, 0)$ , and $B (0, b)$ :
$x = \frac{1 ( 0 ) + 2 ( a )}{1 + 2}$

Solving for Parameter $a$

Simplifying the expression:
$x = \frac{2 a}{3}$
Rearranging to express $a$ in terms of $x$ :
$a = \frac{3 x}{2}$

Section Formula for $y$

Similarly, for the $y$ -coordinate of $P$ :
$y = \frac{m \cdot y _{2} + n \cdot y _{1}}{m + n}$
Substituting the values:
$y = \frac{1 ( b ) + 2 ( 0 )}{1 + 2}$

Solving for Parameter $b$

Simplifying the expression:
$y = \frac{b}{3}$
Rearranging to express $b$ in terms of $y$ :
$b = 3 y$

Eliminating the Parameters

Recall the constraint equation: $a^{2} + b^{2} = l^{2}$
Substitute $a = \frac{3 x}{2}$ and $b = 3 y$ into this equation:
$(\frac{3 x}{2})^{2} + (3 y)^{2} = l^{2}$

Expanding the Squares

Expanding the squared terms:
$\frac{9 x ^{2}}{4} + 9 y^{2} = l^{2}$
Notice how the variables $a$ and $b$ have been completely eliminated.

The Equation of the Locus

To clear the fraction, multiply the entire equation by $4$ :
$9 x^{2} + 36 y^{2} = 4 l^{2}$
This is the equation of an ellipse.

Key Takeaways & Challenge

Locus: The path traced by point $P$ is an ellipse.
General Rule: Any point dividing a sliding segment in a fixed ratio (other than $1 : 1$ ) traces an ellipse.
Challenge: What if $P$ is the midpoint (ratio $1 : 1$ )? Does it trace a circle? Try it out!

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The Sigma Insight: Distance and Section Formulas

Solution Diagram

Analyzing the Setup

Imagine a rod of fixed length

l

leaning against a wall. Let the floor be the

x

-axis and the wall be the

y

-axis, with the endpoints of the rod at

A (a, 0)

and

B (0, b)

As the rod slides, the values of

a

and

b

change, but the length of the rod remains constant. This provides our fundamental constraint based on the Pythagorean theorem:

a^{2} + b^{2} = l^{2}

This equation serves as the heartbeat of our problem, defining the boundary conditions for the motion of the rod.

The Bridge

The Section Formula

We are interested in the path traced by a point

P (x, y)

that divides the rod

A B

in the ratio

1 : 2

. We must express the coordinates of

P

in terms of the variable intercepts

a

and

b

Using the section formula for internal division with a ratio of

m : n = 1 : 2

, we calculate the

x

-coordinate:

x = \frac{1 ( 0 ) + 2 ( a )}{1 + 2} = \frac{2 a}{3}

Similarly, we calculate the

y

-coordinate:

y = \frac{1 ( b ) + 2 ( 0 )}{1 + 2} = \frac{b}{3}

These equations act as our bridge, connecting the position of point

P

to the parameters

a

and

b

The Elimination

Unveiling the Locus

To find the locus of point

P

, we must eliminate the parameters

a

and

b

using our bridge equations and the constraint equation. From the bridge equations, we isolate

a

and

b

a = \frac{3 x}{2}, b = 3 y

Substituting these expressions into the constraint equation

a^{2} + b^{2} = l^{2}

, we obtain:

(\frac{3 x}{2})^{2} + (3 y)^{2} = l^{2}

Expanding these terms, we arrive at:

\frac{9 x ^{2}}{4} + 9 y^{2} = l^{2}

The Final Reveal

To express the path in a more elegant standard form, we multiply the entire equation by

4

9 x^{2} + 36 y^{2} = 4 l^{2}

Dividing by

4 l^{2}

, we can see the standard form of an ellipse:

\frac{x ^{2}}{( 2 l /3 ) ^{2}} + \frac{y ^{2}}{( l /3 ) ^{2}} = 1

We have successfully translated a dynamic physical motion into a static geometric shape. The path traced by point

P

is an ellipse.

Common Traps & FAQs

Wait, why do we use the section formula here? I thought that was only for fixed points.

The section formula is a powerful tool for any point dividing a segment in a fixed ratio, regardless of whether the segment is moving or static. Because the ratio 1:2 is constant, the section formula gives us a direct, algebraic link between the moving endpoints and the point P, allowing us to track P's path as the rod slides.

Why is the locus an ellipse? I expected a straight line.

That is a great intuition check! A straight line would only occur if the point P were moving in a linear fashion. However, because the rod is constrained by both axes simultaneously, the point P is pulled in two directions at once. This 'tug-of-war' between the x and y constraints is exactly what creates the curvature of an ellipse.

What happens if the ratio is 1:1? Does the math still work?

Absolutely! If the ratio is 1:1, P is the midpoint. The section formula would give x = a/2 and y = b/2. Substituting these into a^2 + b^2 = l^2 leads to (2x)^2 + (2y)^2 = l^2, which simplifies to x^2 + y^2 = (l/2)^2. This is the equation of a circle with radius l/2. The ellipse 'collapses' into a circle when the ratio is 1:1!

Animated Solution for Mathematics - Straight Lines: A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

Visualized Solution

Visualizing the Sliding Rod

Coordinates of the Moving Ends

Applying the Length Constraint

Locating the Point $P (x, y)$

Section Formula for $x$

Solving for Parameter $a$

Section Formula for $y$

Solving for Parameter $b$

Eliminating the Parameters

Expanding the Squares

The Equation of the Locus

Key Takeaways & Challenge

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Bridge

The Elimination

The Final Reveal

Similar Questions

If the equation of the locus of a point equidistant from the point $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then the value of 'c' is

If the line $2 x + y = k$ passes through the point which divides the line segment joining the points $(1, 1)$ and $(2, 4)$ in the ratio $3 : 2$ , then $k$ equals

For $a > b > c > 0$ , the distance between $(1, 1)$ and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less than $22$ . Then

Let $S$ be a square of unit area. Consider any quadrilateral which has one vertex on each side of $S$ . If $a, b, c$ , and $d$ denote the lengths of the sides of the quadrilateral, prove that $2 \leq a^{2} + b^{2} + c^{2} + d^{2} \leq 4$ .

Let $A (- 2, - 1), B (1, 0), C (α, β)$ and $D (γ, δ)$ be the vertices of a parallelogram $A B C D$ . If the point $C$ lies on $2 x - y = 5$ and the point $D$ lies on $3 x - 2 y = 6$ , then the value of $∣ α + β + γ + δ ∣$ is equal to _______.

If $(1, 2), (4, 6), (5, 7)$ and $(a, b)$ are the vertices of a parallelogram $P QR S$ , then

A triangle with vertices $(4, 0), (- 1, - 1), (3, 5)$ is

Visualized Solution

Visualizing the Sliding Rod

Coordinates of the Moving Ends

Applying the Length Constraint

Locating the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x,y)

Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Solving for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Solving for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Eliminating the Parameters

Expanding the Squares

The Equation of the Locus

Key Takeaways & Challenge

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Bridge

The Elimination

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the equation of the locus of a point equidistant from the point (a1​,b1​) and (a2​,b2​) is (a1​−a2​)x+(b1​−b2​)y+c=0, then the value of 'c' is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a>b>c>0, the distance between (1,1) and the point of intersection of the lines ax+by+c=0 and bx+ay+c=0 is less than 22​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (1,2),(4,6),(5,7) and (a,b) are the vertices of a parallelogram PQRS, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A triangle with vertices (4,0),(−1,−1),(3,5) is

Locating the Point $P (x, y)$

Section Formula for $x$

Solving for Parameter $a$

Section Formula for $y$

Solving for Parameter $b$

If the equation of the locus of a point equidistant from the point $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then the value of 'c' is

If the line $2 x + y = k$ passes through the point which divides the line segment joining the points $(1, 1)$ and $(2, 4)$ in the ratio $3 : 2$ , then $k$ equals

For $a > b > c > 0$ , the distance between $(1, 1)$ and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less than $22$ . Then

If $(1, 2), (4, 6), (5, 7)$ and $(a, b)$ are the vertices of a parallelogram $P QR S$ , then

A triangle with vertices $(4, 0), (- 1, - 1), (3, 5)$ is