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 $\triangle PAB$ is 10. If the locus of $P$ is $ax + by = 15$, then $5a + 2b$ is : Options: (A) 6 (B) $-\frac{6}{5}$ (C) 4 (D) $-\frac{12}{5}$

### Analyzing the Setup Imagine you are standing on a coordinate plane with two fixed anchors, point $A(-1, 1)$ and point $B(2, 3)$. These two points define a rigid, unmoving line segment. A third point, $P(x, y)$, moves such that the area of the triangle formed by $P$, $A$, and $B$ is always exactly $10$ square units. We must determine the path traced by $P$. ### The Mathematical Foundation To solve this, we use the coordinate geometry formula for the area of a triangle: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Substituting our points $A(-1, 1)$, $B(2, 3)$, and $P(h, k)$, we set up the following equation: $$10 = \frac{1}{2} |(-1)(3 - k) + 2(k - 1) + h(1 - 3)|$$ Simplifying the expression inside the modulus, we obtain: $$20 = |-3 + k + 2k - 2 - 2h|$$ This simplifies to the locus equation: $$|2h - 3k + 5| = 20$$ ### The 'Above the Line' Constraint The modulus operator implies two possible lines: $2h - 3k + 5 = 20$ or $2h - 3k + 5 = -20$. To determine the correct path, we first find the equation of line $AB$. The slope of $AB$ is $m = \frac{3-1}{2-(-1)} = \frac{2}{3}$. Using the point-slope form, we get $y - 1 = \frac{2}{3}(x + 1)$, which simplifies to: $$2x - 3y + 5 = 0$$ For a point $(h, k)$ to be "above" this line, the expression $2h - 3k + 5$ must be negative. Consequently, we must open the modulus with a negative sign: $$-(2h - 3k + 5) = 20$$ This simplifies to the final locus equation: $$2h - 3k = -25$$ ### The Final Transformation We are tasked with matching this result to the form $ax + by = 15$. We multiply the equation $2x - 3y = -25$ by $-\frac{3}{5}$ to align the constant term: $$-\frac{6}{5}x + \frac{9}{5}y = 15$$ By comparing coefficients, we identify $a = -\frac{6}{5}$ and $b = \frac{9}{5}$. We now evaluate the expression $5a + 2b$: $$5\left(-\frac{6}{5}\right) + 2\left(\frac{9}{5}\right) = -6 + \frac{18}{5}$$ Calculating the final value: $$\frac{-30 + 18}{5} = -\frac{12}{5}$$ The final result is **$-\frac{12}{5}$**.

JEE Main 2024 (05 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: 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 :

Q: Wait, why does the locus turn out to be a line? I thought it would be a curve.

That is a fantastic intuition to question! In coordinate geometry, the area of a triangle with a fixed base $AB$ is given by $\frac{1}{2} \times \text{base} \times \text{height}$. Since the base $AB$ is fixed in length and the area is fixed at 10, the height of the triangle must be constant. A point that maintains a constant perpendicular distance from a fixed line traces a path parallel to that line. Hence, the locus is a straight line.

Q: How do I know if the expression inside the modulus is positive or negative?

Great question! The condition 'above the line' is your key. First, find the equation of the line $AB$, which is $2x - 3y + 5 = 0$. If you rewrite this as $y = \frac{2}{3}x + \frac{5}{3}$, any point $(h, k)$ above the line must satisfy $k > \frac{2}{3}h + \frac{5}{3}$. Rearranging this inequality gives $2h - 3k + 5 < 0$. Since the expression is negative, the modulus $|2h - 3k + 5|$ must be replaced by $-(2h - 3k + 5)$.

Q: Why did we multiply the final equation by $-\frac{3}{5}$?

We did that to match the target form $ax + by = 15$. Our derived equation was $2x - 3y = -25$. To turn the constant $-25$ into $15$, we multiplied by the ratio $\frac{15}{-25}$, which is $-\frac{3}{5}$. It is just a matching game to identify the coefficients $a$ and $b$ correctly!

Select Answer:

Visualized Solution

Visualizing the Problem

Points: $A (- 1, 1)$ and $B (2, 3)$
Variable point $P (x, y)$ is above line $A B$
Area of $△ P A B = 10$

The Area Formula

Area of $△ P A B = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$
Let $P = (h, k)$ , $A = (- 1, 1)$ , $B = (2, 3)$

Substituting the Coordinates

$10 = \frac{1}{2} ∣ (- 1) (3 - k) + 2 (k - 1) + h (1 - 3) ∣$

Simplifying the Expression

$20 = ∣ - 3 + k + 2 k - 2 - 2 h ∣$
$20 = ∣3 k - 2 h - 5∣$
$∣2 h - 3 k + 5∣ = 20$

Equation of Line $A B$

Slope of $A B = \frac{3 - 1}{2 - ( - 1 )} = \frac{2}{3}$
Equation of $A B$ : $y - 1 = \frac{2}{3} (x + 1)$
$2 x - 3 y + 5 = 0$

The 'Above the Line' Condition

Line $A B$ : $y = \frac{2}{3} x + \frac{5}{3}$
$P (h, k)$ is above $A B \Rightarrow k > \frac{2}{3} h + \frac{5}{3}$
$3 k > 2 h + 5 \Rightarrow 2 h - 3 k + 5 < 0$

Selecting the Correct Locus

Since $2 h - 3 k + 5 < 0$ , $∣2 h - 3 k + 5∣ = - (2 h - 3 k + 5)$
$- (2 h - 3 k + 5) = 20$
$2 h - 3 k = - 25$
Locus of $P$ : $2 x - 3 y = - 25$

Matching the Standard Form

Given locus: $a x + b y = 15$
Multiply $2 x - 3 y = - 25$ by $- \frac{3}{5}$
$- \frac{6}{5} x + \frac{9}{5} y = 15$

Identifying Constants $a$ and $b$

Comparing $- \frac{6}{5} x + \frac{9}{5} y = 15$ with $a x + b y = 15$ :
$a = - \frac{6}{5}$
$b = \frac{9}{5}$

Final Calculation

Calculate $5 a + 2 b$ :
$5 (- \frac{6}{5}) + 2 (\frac{9}{5})$
$= - 6 + \frac{18}{5}$
$= \frac{- 30 + 18}{5} = - \frac{12}{5}$

Conclusion and Key Takeaway

Key Takeaway:
The condition "above the line" determines the sign of the linear expression in the area formula.
Final Answer: $- \frac{12}{5}$

00:00 / 00:00

The Sigma Insight: Area of Triangle

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane with two fixed anchors, point

A (- 1, 1)

and point

B (2, 3)

. These two points define a rigid, unmoving line segment.

A third point,

P (x, y)

, moves such that the area of the triangle formed by

P

A

, and

B

is always exactly

10

square units. We must determine the path traced by

P

The Mathematical Foundation

To solve this, we use the coordinate geometry formula for the area of a triangle:

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

Substituting our points

A (- 1, 1)

B (2, 3)

, and

P (h, k)

, we set up the following equation:

10 = \frac{1}{2} ∣ (- 1) (3 - k) + 2 (k - 1) + h (1 - 3) ∣

Simplifying the expression inside the modulus, we obtain:

20 = ∣ - 3 + k + 2 k - 2 - 2 h ∣

This simplifies to the locus equation:

∣2 h - 3 k + 5∣ = 20

The 'Above the Line' Constraint

The modulus operator implies two possible lines:

2 h - 3 k + 5 = 20

2 h - 3 k + 5 = - 20

. To determine the correct path, we first find the equation of line

A B

The slope of

A B

m = \frac{3 - 1}{2 - ( - 1 )} = \frac{2}{3}

. Using the point-slope form, we get

y - 1 = \frac{2}{3} (x + 1)

, which simplifies to:

2 x - 3 y + 5 = 0

For a point

(h, k)

to be "above" this line, the expression

2 h - 3 k + 5

must be negative. Consequently, we must open the modulus with a negative sign:

- (2 h - 3 k + 5) = 20

This simplifies to the final locus equation:

2 h - 3 k = - 25

The Final Transformation

We are tasked with matching this result to the form

a x + b y = 15

. We multiply the equation

2 x - 3 y = - 25

- \frac{3}{5}

to align the constant term:

- \frac{6}{5} x + \frac{9}{5} y = 15

By comparing coefficients, we identify

a = - \frac{6}{5}

and

b = \frac{9}{5}

. We now evaluate the expression

5 a + 2 b

5 (- \frac{6}{5}) + 2 (\frac{9}{5}) = - 6 + \frac{18}{5}

Calculating the final value:

\frac{- 30 + 18}{5} = - \frac{12}{5}

The final result is $- \frac{12}{5}$ .

Common Traps & FAQs

Wait, why does the locus turn out to be a line? I thought it would be a curve.

That is a fantastic intuition to question! In coordinate geometry, the area of a triangle with a fixed base

A B

is given by

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

. Since the base

A B

is fixed in length and the area is fixed at 10, the height of the triangle must be constant. A point that maintains a constant perpendicular distance from a fixed line traces a path parallel to that line. Hence, the locus is a straight line.

How do I know if the expression inside the modulus is positive or negative?

Great question! The condition 'above the line' is your key. First, find the equation of the line

A B

, which is

2 x - 3 y + 5 = 0

. If you rewrite this as

y = \frac{2}{3} x + \frac{5}{3}

, any point

(h, k)

above the line must satisfy

k > \frac{2}{3} h + \frac{5}{3}

. Rearranging this inequality gives

2 h - 3 k + 5 < 0

. Since the expression is negative, the modulus

∣2 h - 3 k + 5∣

must be replaced by

- (2 h - 3 k + 5)

Why did we multiply the final equation by

- \frac{3}{5}

We did that to match the target form

a x + b y = 15

. Our derived equation was

2 x - 3 y = - 25

. To turn the constant

- 25

into

15

, we multiplied by the ratio

\frac{15}{- 25}

, which is

- \frac{3}{5}

. It is just a matching game to identify the coefficients

a

and

b

correctly!

Animated Solution for Mathematics - Straight Lines: 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 :

Select Answer:

Visualized Solution

Visualizing the Problem

The Area Formula

Substituting the Coordinates

Simplifying the Expression

Equation of Line $A B$

The 'Above the Line' Condition

Selecting the Correct Locus

Matching the Standard Form

Identifying Constants $a$ and $b$

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Mathematical Foundation

The 'Above the Line' Constraint

The Final Transformation

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 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:

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 $P$ .

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 :

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 $C$ .

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 .

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 $A (4, - 2), B (1, 1)$ and $C (9, - 3)$ be the vertices of a triangle $A B C$ . Then the maximum area of the parallelogram $A F D E$ , formed with vertices $D, E$ and $F$ on the sides $B C, C A$ and $A B$ of the triangle $A B C$ respectively, is ______ .

Select Answer:

Visualized Solution

Visualizing the Problem

The Area Formula

Substituting the Coordinates

Simplifying the Expression

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

The 'Above the Line' Condition

Selecting the Correct Locus

Matching the Standard Form

Identifying Constants .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

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Mathematical Foundation

The 'Above the Line' Constraint

The Final Transformation

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 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; } 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 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; } 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

Equation of Line $A B$

Identifying Constants $a$ and $b$

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 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:

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 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 .

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