JEE Advanced 1988

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Animated Solution for Mathematics - Straight Lines: If $P = (1, 0), Q = (- 1, 0)$ and $R = (2, 0)$ are three given points, then locus of the point $S$ satisfying the relation $S Q^{2} + S R^{2} = 2 S P^{2}$ , is

Q: Wait, why do we use $SQ^2$ instead of just $SQ$? Wouldn't it be more direct to use the distance formula without squaring?

That is a fantastic question! If you used $SQ$ without squaring, you would be dealing with square roots like $\sqrt{(x+1)^2 + y^2}$. Integrating or manipulating those in an equation is a nightmare. By working with the squared distances, we keep our algebra clean, polynomial, and manageable. Always look for ways to simplify the expression before you start calculating!

Q: How do I know if the line is parallel to the x-axis or the y-axis? I always get confused.

Think about the equation $x = -\frac{3}{2}$. This equation says that no matter what value $y$ takes, $x$ must always be $-1.5$. If you plot this, you get a vertical line. Since the y-axis is also vertical, a line where $x$ is constant must be parallel to the y-axis. Conversely, if you had $y = constant$, that would be a horizontal line parallel to the x-axis.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Think of the 'locus' as a trail left by a particle moving in a plane. Before you touch your pen to paper, visualize the three points

P (1, 0)

Q (- 1, 0)

, and

R (2, 0)

sitting on the x-axis. The condition

S Q^{2} + S R^{2} = 2 S P^{2}

is simply a constraint on the particle's freedom. What does this constraint imply about the balance of distances?

Hint 2

Hint 2: You have the distance formula in your toolkit:

d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

. Use this to express

S Q^{2}

S R^{2}

, and

S P^{2}

in terms of the coordinates of

S (x, y)

. Do not worry about square roots; the problem gives you squared distances, which is a massive gift to simplify your algebra.

Hint 3

Hint 3: Write out the full equation:

(x + 1)^{2} + y^{2} + (x - 2)^{2} + y^{2} = 2 [(x - 1)^{2} + y^{2}]

. Now, expand the squares carefully. You will notice that the

x^{2}

and

y^{2}

terms on both sides are suspiciously similar. Don't panic when they cancel out—that is exactly what should happen to reveal the nature of the locus.

Visualized Solution

Plotting the Fixed Points $P$ , $Q$ , and $R$

We are given three fixed points on the x-axis: $P (1, 0)$ , $Q (- 1, 0)$ , and $R (2, 0)$ .
Since their y-coordinates are all $0$ , these points lie entirely on the horizontal axis.

Introducing the Moving Point $S (x, y)$

Let $S (x, y)$ be any arbitrary point in the coordinate plane.
The position of $S$ will vary, but it must always satisfy the given geometric condition.

The Geometric Condition

The given relation is: $S Q^{2} + S R^{2} = 2 S P^{2}$ .
This equation relates the squared distances from $S$ to the three fixed points $Q$ , $R$ , and $P$ .

Finding the Squared Distance $S Q^{2}$

Recall the distance formula: $d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
For points $S (x, y)$ and $Q (- 1, 0)$ :
$S Q^{2} = (x - (- 1))^{2} + (y - 0)^{2} = (x + 1)^{2} + y^{2}$

Finding the Squared Distance $S R^{2}$

For points $S (x, y)$ and $R (2, 0)$ :
$S R^{2} = (x - 2)^{2} + (y - 0)^{2} = (x - 2)^{2} + y^{2}$

Finding the Squared Distance $S P^{2}$

For points $S (x, y)$ and $P (1, 0)$ :
$S P^{2} = (x - 1)^{2} + (y - 0)^{2} = (x - 1)^{2} + y^{2}$

Substituting into the Relation

Substitute the expressions into the given equation:
$(x + 1)^{2} + y^{2} + (x - 2)^{2} + y^{2} = 2 [(x - 1)^{2} + y^{2}]$

Expanding the Left-Hand Side

LHS: $(x + 1)^{2} + y^{2} + (x - 2)^{2} + y^{2}$
$= (x^{2} + 2 x + 1) + y^{2} + (x^{2} - 4 x + 4) + y^{2}$
$= 2 x^{2} - 2 x + 5 + 2 y^{2}$

Expanding the Right-Hand Side

RHS: $2 [(x - 1)^{2} + y^{2}]$
$= 2 [x^{2} - 2 x + 1 + y^{2}]$
$= 2 x^{2} - 4 x + 2 + 2 y^{2}$

Equating and Cancelling Common Terms

Equate LHS and RHS:
$2 x^{2} - 2 x + 5 + 2 y^{2} = 2 x^{2} - 4 x + 2 + 2 y^{2}$
Subtract $2 x^{2} + 2 y^{2}$ from both sides:
$- 2 x + 5 = - 4 x + 2$

Solving the Linear Equation

Rearrange terms to solve for $x$ :
$- 2 x + 4 x = 2 - 5$
$2 x = - 3$
$x = - \frac{3}{2}$

Identifying the Locus Geometrically

The equation of the locus is $x = - \frac{3}{2}$ .
This represents a vertical line parallel to the y-axis.
Therefore, the correct option is (d).

00:00 / 00:00

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

We are given three fixed points on the Cartesian plane:

P (1, 0)

Q (- 1, 0)

, and

R (2, 0)

. These points are collinear, lying entirely on the

x

-axis.

Let the moving point

S

be represented by the coordinates

(x, y)

. The motion of

S

is governed by the geometric constraint:

S Q^{2} + S R^{2} = 2 S P^{2}

The Algebraic Translation

To translate this geometric condition into an algebraic equation, we utilize the distance formula. The squared distance between two points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

is given by

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Applying this to our specific points, we define the squared distances as follows:

S Q^{2} = (x + 1)^{2} + y^{2}

S R^{2} = (x - 2)^{2} + y^{2}

S P^{2} = (x - 1)^{2} + y^{2}

Substituting these expressions into our governing equation, we obtain:

(x + 1)^{2} + y^{2} + (x - 2)^{2} + y^{2} = 2 [(x - 1)^{2} + y^{2}]

The Great Cancellation

Next, we expand both sides of the equation to simplify the expression.

Expanding the Left-Hand Side (LHS):

(x^{2} + 2 x + 1) + y^{2} + (x^{2} - 4 x + 4) + y^{2} = 2 x^{2} - 2 x + 5 + 2 y^{2}

Expanding the Right-Hand Side (RHS):

2 [(x^{2} - 2 x + 1) + y^{2}] = 2 x^{2} - 4 x + 2 + 2 y^{2}

Equating the two sides, we have:

2 x^{2} - 2 x + 5 + 2 y^{2} = 2 x^{2} - 4 x + 2 + 2 y^{2}

Notice that the terms

2 x^{2}

and

2 y^{2}

appear on both sides of the equation. By subtracting

2 x^{2} + 2 y^{2}

from both sides, the quadratic components vanish entirely.

Final Calculation

We are left with a simple linear equation:

- 2 x + 5 = - 4 x + 2

Rearranging the terms to solve for

x

4 x - 2 x = 2 - 5

2 x = - 3

x = - \frac{3}{2}

The locus of the point

S

is the vertical straight line defined by the equation $x = - 1.5$ . This line is parallel to the

y

-axis and represents all possible positions the dancer can occupy while satisfying the given condition.

Common Traps & FAQs

Wait, why do we use

S Q^{2}

instead of just

S Q

? Wouldn't it be more direct to use the distance formula without squaring?

That is a fantastic question! If you used

S Q

without squaring, you would be dealing with square roots like

(x + 1)^{2} + y^{2}

. Integrating or manipulating those in an equation is a nightmare. By working with the squared distances, we keep our algebra clean, polynomial, and manageable. Always look for ways to simplify the expression before you start calculating!

I got to the step where

x^{2}

and

y^{2}

cancel out. Did I make a mistake? Shouldn't a locus be a curve?

Take a deep breath—you did exactly the right thing! In coordinate geometry, a 'curve' can be a straight line. When the quadratic terms cancel out, it simply tells us that the locus is not a circle or a parabola, but a straight line. It is a beautiful moment of simplification, not an error.

How do I know if the line is parallel to the x-axis or the y-axis? I always get confused.

Think about the equation

x = - \frac{3}{2}

. This equation says that no matter what value

y

takes,

x

must always be

- 1.5

. If you plot this, you get a vertical line. Since the y-axis is also vertical, a line where

x

is constant must be parallel to the y-axis. Conversely, if you had

y = co n s t an t

, that would be a horizontal line parallel to the x-axis.

Animated Solution for Mathematics - Straight Lines: If $P = (1, 0), Q = (- 1, 0)$ and $R = (2, 0)$ are three given points, then locus of the point $S$ satisfying the relation $S Q^{2} + S R^{2} = 2 S P^{2}$ , is

Select Answer:

Visualized Solution

Plotting the Fixed Points $P$ , $Q$ , and $R$

Introducing the Moving Point $S (x, y)$

The Geometric Condition

Finding the Squared Distance $S Q^{2}$

Finding the Squared Distance $S R^{2}$

Finding the Squared Distance $S P^{2}$

Substituting into the Relation

Expanding the Left-Hand Side

Expanding the Right-Hand Side

Equating and Cancelling Common Terms

Solving the Linear Equation

Identifying the Locus Geometrically

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Algebraic Translation

The Great Cancellation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

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

The Geometric Condition

Finding the Squared Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SQ2

Finding the Squared Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SR2

Finding the Squared Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SP2

Substituting into the Relation

Expanding the Left-Hand Side

Expanding the Right-Hand Side

Equating and Cancelling Common Terms

Solving the Linear Equation

Identifying the Locus Geometrically

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Algebraic Translation

The Great Cancellation

Final Calculation

Similar Questions

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Introducing the Moving Point $S (x, y)$

Finding the Squared Distance $S Q^{2}$

Finding the Squared Distance $S R^{2}$

Finding the Squared Distance $S P^{2}$

The locus of the mid-point of the perpendiculars drawn from points on the line, $x = 2 y$ to the line $x = y$ is:

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

Locus of mid point of the portion between the axes of $x cos α + y sin α = p$ where $p$ is constant is

The locus of mid points of the perpendiculars drawn from points on the line $x = 2 y$ to the line $x = y$ is :

A rectangle $P QR S$ has its side $P Q$ parallel to the line $y = m x$ and vertices $P, Q$ and $S$ on the lines $y = a, x = b$ and $x = - b$ , respectively. Find the locus of the vertex $R$ .