JEE Advanced 1996

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: 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 $R$ .

Visualized Solution

Visualizing the Constraints

Rectangle $P QR S$ with $P Q ∥ y = m x$ .
Vertices constraints:
Vertex $P$ lies on $y = a$ .
Vertex $Q$ lies on $x = b$ .
Vertex $S$ lies on $x = - b$ .

Assigning Coordinates

Let $P = (x_{1}, a)$
Let $Q = (b, y_{1})$
Let $S = (- b, y_{2})$
Let $R = (x, y)$

Slope of $P Q$

Slope of $P Q = m$
$\frac{y _{1} - a}{b - x _{1}} = m$
Equation 1: $y_{1} - a = m (b - x_{1})$

Slope of $P S$

$P Q ⊥ P S ⟹$ Slope of $P S = - \frac{1}{m}$
$\frac{y _{2} - a}{- b - x _{1}} = - \frac{1}{m}$
Equation 2: $m (y_{2} - a) = b + x_{1}$

Vector Equality for $x$

In a rectangle, $P Q = S R$
Comparing x-components: $b - x_{1} = x - (- b)$
$b - x_{1} = x + b$
$x_{1} = - x$

Vector Equality for $y$

Comparing y-components: $y_{1} - a = y - y_{2}$
$y_{2} = y - y_{1} + a$

Substituting $x_{1}$ into Equation 1

Substitute $x_{1} = - x$ into Equation 1:
$y_{1} - a = m (b - (- x))$
$y_{1} - a = m (b + x)$
$y_{1} = a + m (b + x)$

Substituting into Equation 2

Substitute $x_{1} = - x$ and $y_{2} = y - y_{1} + a$ into Equation 2:
$m ((y - y_{1} + a) - a) = b + (- x)$
$m (y - y_{1}) = b - x$

Eliminating $y_{1}$

Substitute $y_{1} = a + m (b + x)$ into $m (y - y_{1}) = b - x$ :
$m [y - (a + m (b + x))] = b - x$

Algebraic Expansion

Expanding the brackets:
$m y - ma - m^{2} (b + x) = b - x$
$m y - ma - m^{2} b - m^{2} x = b - x$

The Final Locus Equation

Rearranging terms to one side:
$m^{2} x - x - m y + m^{2} b + b + am = 0$
Final Equation:
$x (m^{2} - 1) - y m + (m^{2} + 1) b + am = 0$

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The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Setup

We are tracking the vertex

R (x, y)

of a rectangle

P QR S

. The vertices are constrained as follows:

P

lies on

y = a

Q

lies on

x = b

, and

S

lies on

x = - b

Let the coordinates be

P = (x_{1}, a)

Q = (b, y_{1})

, and

S = (- b, y_{2})

. The side

P Q

is parallel to the line

y = m x

, which implies the slope of

P Q

m

Establishing Geometric Constraints

Using the slope formula for

P Q

, we have:

\frac{y _{1} - a}{b - x _{1}} = m ⟹ y_{1} - a = m (b - x_{1})

Since

P Q ⊥ P S

in a rectangle, the slope of

P S

must be

- \frac{1}{m}

. Applying this to the coordinates of

P

and

S

\frac{y _{2} - a}{- b - x _{1}} = - \frac{1}{m} ⟹ m (y_{2} - a) = b + x_{1}

Utilizing Vector Equality

In any rectangle, the vector

P Q

must equal the vector

S R

. By equating the

x

-components:

b - x_{1} = x - (- b) ⟹ b - x_{1} = x + b ⟹ x_{1} = - x

Equating the

y

-components gives:

y_{1} - a = y - y_{2} ⟹ y_{2} = y - y_{1} + a

The Master Equation

We now substitute

x_{1} = - x

into the slope equations. From the first equation:

y_{1} = a + m (b + x)

Substituting

x_{1} = - x

and

y_{2} = y - y_{1} + a

into the second slope equation

m (y_{2} - a) = b + x_{1}

m (y - y_{1}) = b - x

Final Calculation

To find the locus, we eliminate

y_{1}

by substituting its expression in terms of

x

m [y - (a + m (b + x))] = b - x

Expanding the terms:

m y - ma - m^{2} (b + x) = b - x

m y - ma - m^{2} b - m^{2} x = b - x

Rearranging all terms to one side, we arrive at the final equation for the locus of

R

$x (m^{2} - 1) - y m + (m^{2} + 1) b + am = 0$

Common Traps & FAQs

Wait, why do we use vector equality

P Q = S R

instead of just using the distance formula or midpoint properties?

Great question! While you *could* use midpoints (the diagonals of a rectangle bisect each other), vector equality is often cleaner in coordinate geometry. It directly relates the components of the sides, allowing us to equate the

x

and

y

displacements separately. It turns a geometric problem into a system of linear equations, which is much harder to mess up!

I'm confused about the slope of

P S

. Why is it

- 1/ m

Remember the fundamental property of a rectangle: adjacent sides are perpendicular. If the slope of line

P Q

m

, any line perpendicular to it must have a slope that is the negative reciprocal,

- 1/ m

. This is a classic JEE concept—never forget it!

Why do we reject the negative root or worry about signs? Is there a trap here?

In this specific locus problem, there are no quadratic roots to reject because the equations are linear in terms of the coordinates. The 'trap' isn't in the algebra, but in the setup. If you misplace a negative sign when calculating the slope of

P S

(e.g., forgetting the

- b - x_{1}

in the denominator), the entire locus will shift. Always double-check your coordinate differences!

Animated Solution for Mathematics - Straight Lines: 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 $R$ .

Visualized Solution

Visualizing the Constraints

Assigning Coordinates

Slope of $P Q$

Slope of $P S$

Vector Equality for $x$

Vector Equality for $y$

Substituting $x_{1}$ into Equation 1

Substituting into Equation 2

Eliminating $y_{1}$

Algebraic Expansion

The Final Locus Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Establishing Geometric Constraints

Utilizing Vector Equality

The Master Equation

Final Calculation

Similar Questions

A line cuts the x-axis at $A (7, 0)$ and the y-axis at $B (0, - 5)$ . A variable line $P Q$ is drawn perpendicular to $A B$ cutting the x-axis in $P$ and the y-axis in $Q$ . If $A Q$ and $B P$ intersect at $R$ , find the locus of $R$ .

A straight line through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q$ . If $O$ is the origin and the rectangle $O P R Q$ is completed, then the locus of $R$ is :

One side of a rectangle lies along the line $4 x + 7 y + 5 = 0$ . Two of its vertices are $(- 3, 1)$ and $(1, 1)$ . Find the equations of the other three sides.

The end $A, B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $O X, O Y$ respectively. If the rectangle $O A P B$ be completed, then show that the locus of the foot of the perpendicular drawn from $P$ to $A B$ is $x^{2/3} + y^{2/3} = c^{2/3}$ .

Let $A$ be a fixed point $(0, 6)$ and $B$ be a moving point $(2 t, 0)$ . Let $M$ be the mid-point of $A B$ and the perpendicular bisector of $A B$ meets the $y$ -axis at $C$ . The locus of the mid-point $P$ of $M C$ 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 $P S$ be the median of the triangle with vertices $P (2, 2), Q (6, - 1)$ and $R (7, 3)$ . The equation of the line passing through $(1, - 1)$ and parallel to $P S$ is

Two sides of a rhombus are along the lines, $x - y + 1 = 0$ and $7 x - y - 5 = 0$ . If its diagonals intersect at $(1, 2)$ , then which one of the following is a vertex of this rhombus?

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

Visualized Solution

Visualizing the Constraints

Assigning Coordinates

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PS

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

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

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​ into Equation 1

Substituting into Equation 2

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​

Algebraic Expansion

The Final Locus Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Establishing Geometric Constraints

Utilizing Vector Equality

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } One side of a rectangle lies along the line 4x+7y+5=0. Two of its vertices are (−3,1) and (1,1). Find the equations of the other three sides.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. The equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let PS be the median of the triangle with vertices P(2,2),Q(6,−1) and R(7,3). The equation of the line passing through (1,−1) and parallel to PS is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two sides of a rhombus are along the lines, x−y+1=0 and 7x−y−5=0. If its diagonals intersect at (1,2), then which one of the following is a vertex of this rhombus?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. Then the equation of the bisector of the angle PQR is

Slope of $P Q$

Slope of $P S$

Vector Equality for $x$

Vector Equality for $y$

Substituting $x_{1}$ into Equation 1

Eliminating $y_{1}$

One side of a rectangle lies along the line $4 x + 7 y + 5 = 0$ . Two of its vertices are $(- 3, 1)$ and $(1, 1)$ . Find the equations of the other three sides.

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 $P S$ be the median of the triangle with vertices $P (2, 2), Q (6, - 1)$ and $R (7, 3)$ . The equation of the line passing through $(1, - 1)$ and parallel to $P S$ is

Two sides of a rhombus are along the lines, $x - y + 1 = 0$ and $7 x - y - 5 = 0$ . If its diagonals intersect at $(1, 2)$ , then which one of the following is a vertex of this rhombus?

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