Let two points be $A(1, -1)$ and $B(0, 2)$. If a point $P(x', y')$ be such that the area of $\Delta PAB = 5$ square units and it lies on the line, $3x + y - 4\lambda = 0$, then a value of $\lambda$ is: Options: (A) 3 (B) -3 (C) 4 (D) 2

### Analyzing the Setup We are given two fixed points, $A(1, -1)$ and $B(0, 2)$. A point $P(x', y')$ moves such that the area of $\triangle PAB$ is constant at $5$ square units. The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ ### The Master Equation Substituting the coordinates of $A(1, -1)$, $B(0, 2)$, and $P(x', y')$ into the area formula, we obtain: $$\frac{1}{2} |1(2 - y') + 0(y' - (-1)) + x'(-1 - 2)| = 5$$ Simplifying the expression inside the absolute value: $$\frac{1}{2} |2 - y' - 3x'| = 5$$ Multiplying both sides by $2$ yields: $$|2 - y' - 3x'| = 10$$ This is equivalent to the expression: $$|3x' + y' - 2| = 10$$ ### Solving for the Locus The modulus equation $|3x' + y' - 2| = 10$ represents two distinct parallel lines: 1. $3x' + y' - 2 = 10 \Rightarrow 3x' + y' = 12$ 2. $3x' + y' - 2 = -10 \Rightarrow 3x' + y' = -8$ The problem states that point $P$ lies on the line $3x + y = 4\lambda$. For this to be true, the line $3x + y = 4\lambda$ must coincide with one of the two lines derived above. ### Final Calculation By comparing $3x + y = 4\lambda$ with our two derived equations, we set $4\lambda$ equal to the constants: Case 1: $4\lambda = 12 \Rightarrow \lambda = 3$ Case 2: $4\lambda = -8 \Rightarrow \lambda = -2$ Given the context of the problem, the valid value for $\lambda$ is **$3$**. This confirms that the locus of a point forming a triangle of constant area with a fixed base is a pair of lines parallel to that base.

JEE Main 2020 - 8 Jan (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: 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:

Select Answer:

Visualized Solution

Visualizing the Points $A$ and $B$

Given fixed points: $A (1, - 1)$ and $B (0, 2)$ .
A variable point $P (x^{'}, y^{'})$ moves such that the area of $Δ P A B$ is exactly $5$ square units.

The Area of a Triangle Formula

The area of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ is given by:
$Area = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$

Substituting the Coordinates

Let $(x_{1}, y_{1}) = (1, - 1)$
Let $(x_{2}, y_{2}) = (0, 2)$
Let $(x_{3}, y_{3}) = (x^{'}, y^{'})$
$\frac{1}{2} ∣1 (2 - y^{'}) + 0 (y^{'} - (- 1)) + x^{'} (- 1 - 2) ∣ = 5$

Simplifying the Expression

Simplifying the terms inside the modulus:
$\frac{1}{2} ∣2 - y^{'} + 0 - 3 x^{'} ∣ = 5$
$∣2 - y^{'} - 3 x^{'} ∣ = 10$
$∣3 x^{'} + y^{'} - 2∣ = 10$

Solving the Modulus Equation

Removing the absolute value yields two possible cases:
Case 1: $3 x^{'} + y^{'} - 2 = 10 ⟹ 3 x^{'} + y^{'} = 12$
Case 2: $3 x^{'} + y^{'} - 2 = - 10 ⟹ 3 x^{'} + y^{'} = - 8$

Comparing with the Given Line

The problem states that point $P$ lies on the line:
$3 x + y - 4 λ = 0 ⟹ 3 x + y = 4 λ$
This line must be one of the locus lines we just found.

Calculating the Value of $λ$

Comparing $3 x + y = 4 λ$ with our two cases:
From Case 1: $4 λ = 12 ⟹ λ = 3$
From Case 2: $4 λ = - 8 ⟹ λ = - 2$

Final Conclusion

Possible values for $λ$ are $3$ and $- 2$ .
Checking the given options: $3$ , $- 3$ , $4$ , $2$ .
The correct value present in the options is $3$ .

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The Sigma Insight: Area of Triangle

Solution Diagram

Analyzing the Setup

We are given two fixed points,

A (1, - 1)

and

B (0, 2)

. A point

P (x^{'}, y^{'})

moves such that the area of

△ P A B

is constant at

5

square units.

The area of a triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

is given by the formula:

Area = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣

The Master Equation

Substituting the coordinates of

A (1, - 1)

B (0, 2)

, and

P (x^{'}, y^{'})

into the area formula, we obtain:

\frac{1}{2} ∣1 (2 - y^{'}) + 0 (y^{'} - (- 1)) + x^{'} (- 1 - 2) ∣ = 5

Simplifying the expression inside the absolute value:

\frac{1}{2} ∣2 - y^{'} - 3 x^{'} ∣ = 5

Multiplying both sides by

2

yields:

∣2 - y^{'} - 3 x^{'} ∣ = 10

This is equivalent to the expression:

∣3 x^{'} + y^{'} - 2∣ = 10

Solving for the Locus

The modulus equation

∣3 x^{'} + y^{'} - 2∣ = 10

represents two distinct parallel lines:

3 x^{'} + y^{'} - 2 = 10 \Rightarrow 3 x^{'} + y^{'} = 12

3 x^{'} + y^{'} - 2 = - 10 \Rightarrow 3 x^{'} + y^{'} = - 8

The problem states that point

P

lies on the line

3 x + y = 4 λ

. For this to be true, the line

3 x + y = 4 λ

must coincide with one of the two lines derived above.

Final Calculation

By comparing

3 x + y = 4 λ

with our two derived equations, we set

4 λ

equal to the constants:

Case 1:

4 λ = 12 \Rightarrow λ = 3

Case 2:

4 λ = - 8 \Rightarrow λ = - 2

Given the context of the problem, the valid value for

λ

is $3$ . This confirms that the locus of a point forming a triangle of constant area with a fixed base is a pair of lines parallel to that base.

Common Traps & FAQs

Why are there two possible lines for the locus of P? Shouldn't there be just one?

The area of a triangle is defined as 0.5 * base * height. Since the base AB is fixed, the height must be constant. However, the point P can be at a distance 'h' on either side of the line AB. Thus, the locus consists of two parallel lines, one on each side of AB.

Can I use the determinant form for the area?

Absolutely! The determinant form is just a more compact way of writing the coordinate formula. It is mathematically equivalent and often preferred for its elegance.

Why did we equate the constant terms?

Once we have the equation of the locus in the form 3x + y + C = 0, and we are given that P lies on 3x + y - 4\lambda = 0, the lines must be identical. Therefore, the constant terms must match.

Animated Solution for Mathematics - Straight Lines: 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:

Select Answer:

Visualized Solution

Visualizing the Points $A$ and $B$

The Area of a Triangle Formula

Substituting the Coordinates

Simplifying the Expression

Solving the Modulus Equation

Comparing with the Given Line

Calculating the Value of $λ$

Final Conclusion

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Master Equation

Solving for the Locus

Final Calculation

Similar Questions

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 $A (- 1, 1)$ 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 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 :

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

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 $P$ .

The coordinates of $A, B, C$ are $(6, 3)$ , $(- 3, 5)$ , $(4, - 2)$ respectively, and $P$ is any point $(x, y)$ . Show that the ratio of the area of the triangles $△ P B C$ and $△ A B C$ is $∣ \frac{x + y - 2}{7} ∣$ .

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 $A (h, k), B (1, 1)$ and $C (2, 1)$ be the vertices of a right angled triangle with $A C$ as its hypotenuse. If the area of the triangle is 1 square unit, then the set of values which 'k' can take is given by

Select Answer:

Visualized Solution

Visualizing the Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

The Area of a Triangle Formula

Substituting the Coordinates

Simplifying the Expression

Solving the Modulus Equation

Comparing with the Given Line

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Final Conclusion

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Master Equation

Solving for the Locus

Final Calculation

Similar Questions

.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 :

.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 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 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.

Visualizing the Points $A$ and $B$

Calculating the Value of $λ$

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 $A (- 1, 1)$ 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 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 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 $C$ .