JEE Advanced 2005

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: The area of the triangle formed by the intersection of a line parallel to x-axis and passing through $P (h, k)$ with the lines $y = x$ and $x + y = 2$ is $4 h^{2}$ . Find the locus of the point $P$ .

Q: Why do we get two equations for the locus?

When we equate the area to $4h^2$, we get $(1 - k)^2 = 4h^2$. When taking the square root of both sides, we must account for both the positive and negative roots: $1 - k = \pm 2h$. This $\pm$ sign indicates that there are two distinct geometric paths (or lines) that satisfy the area condition, which is why we end up with two equations for the locus.

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The line passing through

P (h, k)

is parallel to the x-axis, meaning it is a horizontal line with the equation

y = k

. This line intersects the two fixed lines,

y = x

and

x + y = 2

, creating a triangle. Can you identify the three vertices of this triangle in terms of

k

Hint 2

Hint 2: To find the area, you need the base and the height. The base

B C

lies on the horizontal line

y = k

. Its length is the absolute difference between the x-coordinates of the intersection points. The height is the perpendicular distance from the vertex

A

(where the two fixed lines meet) to the line

y = k

. Use the formula

Area = \frac{1}{2} \times base \times height

Hint 3

Hint 3: Calculate the coordinates of the vertices:

A (1, 1)

B (k, k)

, and

C (2 - k, k)

. The base length is

∣ (2 - k) - k ∣ = 2∣1 - k ∣

. The height is

∣1 - k ∣

. Set the area

\frac{1}{2} \times (2∣1 - k ∣) \times ∣1 - k ∣

equal to

4 h^{2}

and solve for the relationship between

h

and

k

to find the locus.

Visualized Solution

Given Lines

The problem involves two fixed lines: $y = x$ and $x + y = 2$ .
Let's plot these on the coordinate plane.

The Moving Line

A point $P$ has coordinates $(h, k)$ .
A line passes through $P$ and is parallel to the $x$ -axis.
The equation of this horizontal line is $y = k$ .

Finding Vertex $A$

Let's find the intersection of $y = x$ and $x + y = 2$ .
Substitute $y = x$ : $x + x = 2 ⟹ 2 x = 2 ⟹ x = 1$ .
Since $y = x$ , $y = 1$ .
The first vertex is $A (1, 1)$ .

Finding Vertex $B$

Next, find the intersection of $y = k$ and $y = x$ .
Since $y = k$ , substituting this gives $x = k$ .
The second vertex is $B (k, k)$ .

Finding Vertex $C$

Finally, find the intersection of $y = k$ and $x + y = 2$ .
Substitute $y = k$ : $x + k = 2 ⟹ x = 2 - k$ .
The third vertex is $C (2 - k, k)$ .

The Formed Triangle

The vertices of the triangle are $A (1, 1)$ , $B (k, k)$ , and $C (2 - k, k)$ .
The base $B C$ lies on the horizontal line $y = k$ .

Length of Base $B C$

Since $B$ and $C$ have the same $y$ -coordinate, the distance is horizontal.
$B C = ∣ x_{C} - x_{B} ∣ = ∣ (2 - k) - k ∣$ .
$B C = ∣2 - 2 k ∣ = 2∣1 - k ∣$ .

Height of the Triangle

The height is the perpendicular distance from vertex $A (1, 1)$ to the base $y = k$ .
$Height = ∣ y_{A} - y_{base} ∣ = ∣1 - k ∣$ .

Calculating the Area

$Area = \frac{1}{2} \times Base \times Height$ .
$Area = \frac{1}{2} \times (2∣1 - k ∣) \times ∣1 - k ∣$ .
$Area = ∣1 - k ∣^{2} = (1 - k)^{2}$ .

Equating to Given Area

The problem states the area is $4 h^{2}$ .
So, $(1 - k)^{2} = 4 h^{2}$ .
Taking the square root on both sides: $1 - k = \pm 2 h$ .

Finding the Locus

We have $1 - k = 2 h$ or $1 - k = - 2 h$ .
Rearranging for $k$ : $k = 1 - 2 h$ or $k = 1 + 2 h$ .
To find the locus of $P (h, k)$ , replace $(h, k)$ with $(x, y)$ .
The locus is $y = 1 - 2 x$ or $y = 1 + 2 x$ .

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The Sigma Insight: Area of Triangle

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate grid. You have two fixed paths: the line

y = x

, which cuts the plane at a perfect 45-degree angle, and the line

x + y = 2

, which slopes downwards.

These two lines are our anchors. They meet at a specific point, which we shall call vertex

A

. By solving the system

y = x

and

x + y = 2

, we find that

2 x = 2

, so

x = 1

, and consequently

y = 1

Our vertex

A

is firmly planted at

(1, 1)

Defining the Moving Point and Intersections

Now, we introduce a moving point

P (h, k)

. Through this point, we draw a line parallel to the x-axis. This is our 'moving' line, and its equation is simply

y = k

This line is the heartbeat of our triangle. It slices through our two fixed lines, creating two new intersection points,

B

and

C

To find

B

, we intersect

y = k

with

y = x

, which gives us the point

(k, k)

. To find

C

, we intersect

y = k

with

x + y = 2

, which gives us

x + k = 2

, or

x = 2 - k

. So,

C

is at

(2 - k, k)

Calculating the Area

Because the line

y = k

is horizontal, the side

B C

is horizontal. The length of the base

B C

is simply the difference in x-coordinates:

∣ (2 - k) - k ∣ = ∣2 - 2 k ∣ = 2∣1 - k ∣

The height of the triangle is the vertical distance from

A (1, 1)

to the line

y = k

, which is

∣1 - k ∣

. The area of a triangle is given by the formula:

Area = \frac{1}{2} \times base \times height

Substituting our values, we get:

Area = \frac{1}{2} \times (2∣1 - k ∣) \times ∣1 - k ∣ = ∣1 - k ∣^{2} = (1 - k)^{2}

Determining the Locus

The problem states that this area is equal to

4 h^{2}

. So, we set:

(1 - k)^{2} = 4 h^{2}

Taking the square root of both sides, we get:

1 - k = \pm 2 h

Rearranging for

k

, we find

k = 1 - 2 h

k = 1 + 2 h

. Replacing

h

and

k

with

x

and

y

to define the locus, we arrive at our final answer:

$y = 1 - 2 x$ or $y = 1 + 2 x$

Common Traps & FAQs

Wait, why is the base

B C

horizontal?

That is a fantastic observation! The problem states that the line passing through

P (h, k)

is parallel to the x-axis. In coordinate geometry, any line parallel to the x-axis has the equation

y = constant

. Since both vertices

B

and

C

lie on this line, their y-coordinates are identical, which makes the segment

B C

perfectly horizontal. This simplifies our area calculation significantly!

Why do we use

∣1 - k ∣

for the height?

The height of a triangle is the perpendicular distance from a vertex to the opposite side. Here, the opposite side is the horizontal line

y = k

. The vertex

A

is at

(1, 1)

. The perpendicular distance from a point

(x_{1}, y_{1})

to a horizontal line

y = k

is simply the absolute difference of the y-coordinates:

∣ y_{1} - k ∣

. Thus, we get

∣1 - k ∣

Why do we get two equations for the locus?

When we equate the area to

4 h^{2}

, we get

(1 - k)^{2} = 4 h^{2}

. When taking the square root of both sides, we must account for both the positive and negative roots:

1 - k = \pm 2 h

. This

\pm

sign indicates that there are two distinct geometric paths (or lines) that satisfy the area condition, which is why we end up with two equations for the locus.

Animated Solution for Mathematics - Straight Lines: The area of the triangle formed by the intersection of a line parallel to x-axis and passing through $P (h, k)$ with the lines $y = x$ and $x + y = 2$ is $4 h^{2}$ . Find the locus of the point $P$ .

Visualized Solution

Given Lines

The Moving Line

Finding Vertex $A$

Finding Vertex $B$

Finding Vertex $C$

The Formed Triangle

Length of Base $B C$

Height of the Triangle

Calculating the Area

Equating to Given Area

Finding the Locus

The Sigma Insight: Area of Triangle

Analyzing the Setup

Defining the Moving Point and Intersections

Calculating the Area

Determining the Locus

Similar Questions

Let $A (- 1, 1)$ and $B (2, 3)$ be two points and $P$ be a variable point above the line $A B$ such that the area of $△ P A B$ is 10. If the locus of $P$ is $a x + b y = 15$ , then $5 a + 2 b$ is :

Let the points of intersections of the lines $x - y + 1 = 0$ , $x - 2 y + 3 = 0$ and $2 x - 5 y + 11 = 0$ are the mid points of the sides of a triangle $A B C$ . Then the area of the triangle $A B C$ is .

The area of a triangle is 5. Two of its vertices are $A (2, 1)$ and $B (3, - 2)$ . The third vertex $C$ lies on $y = x + 3$ . Find $C$ .

Let the area of the triangle with vertices $A (1, α)$ , $B (α, 0)$ and $C (0, α)$ be 4 sq. units. If the point $(α, - α)$ , $(- α, α)$ and $(α^{2}, β)$ are collinear, then $β$ is equal to

Let two points be $A (1, - 1)$ and $B (0, 2)$ . If a point $P (x^{'}, y^{'})$ be such that the area of $Δ P A B = 5$ square units and it lies on the line, $3 x + y - 4 λ = 0$ , then a value of $λ$ is:

Let two points be $A (1, - 1)$ and $B (0, 2)$ . If a point $P (x^{'}, y^{'})$ be such that the area of $Δ P A B = 5$ sq. units and it lies on the line, $3 x + y - 4 λ = 0$ , then the value of $λ$ is :

Let A(-1,1), B(3, 4) and C(2,0) be given three points. A line $y = m x, m > 0$ , intersects lines AC and BC at point P and Q respectively. Let $A_{1}$ and $A_{2}$ be the areas of $△ A B C$ and $△ P QC$ respectively, such that $A_{1} = 3 A_{2}$ , then the value of m is equal to :

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Visualized Solution

Given Lines

The Moving Line

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Formed Triangle

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

Height of the Triangle

Calculating the Area

Equating to Given Area

Finding the Locus

The Sigma Insight: Area of Triangle

Analyzing the Setup

Defining the Moving Point and Intersections

Calculating the Area

Determining the Locus

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A(−1,1) and B(2,3) be two points and P be a variable point above the line AB such that the area of △PAB is 10. If the locus of P is ax+by=15, then 5a+2b is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the points of intersections of the lines x−y+1=0, x−2y+3=0 and 2x−5y+11=0 are the mid points of the sides of a triangle ABC. Then the area of the triangle ABC is .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of a triangle is 5. Two of its vertices are A(2,1) and B(3,−2). The third vertex C lies on y=x+3. Find C.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the area of the triangle with vertices A(1,α), B(α,0) and C(0,α) be 4 sq. units. If the point (α,−α), (−α,α) and (α2,β) are collinear, then β is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let two points be A(1,−1) and B(0,2). If a point P(x′,y′) be such that the area of ΔPAB=5 square units and it lies on the line, 3x+y−4λ=0, then a value of λ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let two points be A(1,−1) and B(0,2). If a point P(x′,y′) be such that the area of ΔPAB=5 sq. units and it lies on the line, 3x+y−4λ=0, then the value of λ is :

Finding Vertex $A$

Finding Vertex $B$

Finding Vertex $C$

Length of Base $B C$

Let $A (- 1, 1)$ and $B (2, 3)$ be two points and $P$ be a variable point above the line $A B$ such that the area of $△ P A B$ is 10. If the locus of $P$ is $a x + b y = 15$ , then $5 a + 2 b$ is :

Let the points of intersections of the lines $x - y + 1 = 0$ , $x - 2 y + 3 = 0$ and $2 x - 5 y + 11 = 0$ are the mid points of the sides of a triangle $A B C$ . Then the area of the triangle $A B C$ is .

The area of a triangle is 5. Two of its vertices are $A (2, 1)$ and $B (3, - 2)$ . The third vertex $C$ lies on $y = x + 3$ . Find $C$ .

Let the area of the triangle with vertices $A (1, α)$ , $B (α, 0)$ and $C (0, α)$ be 4 sq. units. If the point $(α, - α)$ , $(- α, α)$ and $(α^{2}, β)$ are collinear, then $β$ is equal to

Let two points be $A (1, - 1)$ and $B (0, 2)$ . If a point $P (x^{'}, y^{'})$ be such that the area of $Δ P A B = 5$ square units and it lies on the line, $3 x + y - 4 λ = 0$ , then a value of $λ$ is:

Let two points be $A (1, - 1)$ and $B (0, 2)$ . If a point $P (x^{'}, y^{'})$ be such that the area of $Δ P A B = 5$ sq. units and it lies on the line, $3 x + y - 4 λ = 0$ , then the value of $λ$ is :