JEE Advanced 1978

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

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. You have two fixed points, A and B, and a line

y = x + 3

where the third vertex C must reside. Since C is constrained to this line, you can express its coordinates using a single parameter, say

λ

. If the x-coordinate is

λ

, what must the y-coordinate be based on the line equation?

Hint 2

Hint 2: Recall the coordinate geometry formula for the area of a triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

. The area is given by

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

. Why is the absolute value sign essential here, and what does it imply about the possible locations of vertex C?

Hint 3

Hint 3: Substitute the coordinates

A (2, 1)

B (3, - 2)

, and

C (λ, λ + 3)

into the area formula and set it equal to 5. Your starting equation will be

\frac{1}{2} ∣2 (- 2 - (λ + 3)) + 3 ((λ + 3) - 1) + λ (1 - (- 2)) ∣ = 5

. Solve this absolute value equation for

λ

by considering both the positive and negative cases.

Visualized Solution

Visualizing the Given Data

Given vertices: $A (2, 1)$ and $B (3, - 2)$
Area of $△ A B C = 5$
Vertex $C$ lies on the line: $y = x + 3$

Defining Vertex $C$ Parametrically

Let the $x$ -coordinate of $C$ be $λ$ .
Since $C$ lies on $y = x + 3$ , its $y$ -coordinate is $λ + 3$ .
Coordinates of $C = (λ, λ + 3)$

The Area Formula

Area $= \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$
Substitute Area $= 5$ and coordinates of $A, B, C$ .

Substituting the Values

$\frac{1}{2} ∣2 (- 2 - (λ + 3)) + 3 ((λ + 3) - 1) + λ (1 - (- 2)) ∣ = 5$
Multiplying by $2$ : $∣2 (- 2 - λ - 3) + 3 (λ + 2) + λ (3) ∣ = 10$

Simplifying the Terms

First term: $2 (- 2 - λ - 3) = 2 (- λ - 5) = - 2 λ - 10$
Second term: $3 (λ + 3 - 1) = 3 (λ + 2) = 3 λ + 6$
Third term: $λ (1 - (- 2)) = λ (3) = 3 λ$

Combining the Terms

Summing terms: $(- 2 λ - 10) + (3 λ + 6) + 3 λ$
Grouping $λ$ : $(- 2 + 3 + 3) λ = 4 λ$
Grouping constants: $- 10 + 6 = - 4$
Equation: $∣4 λ - 4∣ = 10$

Handling the Absolute Value

Property: $∣ x ∣ = a \Rightarrow x = a or x = - a$
Case 1: $4 λ - 4 = 10$
Case 2: $4 λ - 4 = - 10$

Solving Case 1

Case 1: $4 λ - 4 = 10 \Rightarrow 4 λ = 14 \Rightarrow λ = \frac{7}{2}$
$x = \frac{7}{2}$ , $y = \frac{7}{2} + 3 = \frac{13}{2}$
Point $C_{1} = (\frac{7}{2}, \frac{13}{2})$

Solving Case 2

Case 2: $4 λ - 4 = - 10 \Rightarrow 4 λ = - 6 \Rightarrow λ = - \frac{3}{2}$
$x = - \frac{3}{2}$ , $y = - \frac{3}{2} + 3 = \frac{3}{2}$
Point $C_{2} = (- \frac{3}{2}, \frac{3}{2})$

Final Conclusion

The two possible coordinates for vertex $C$ are: $(\frac{7}{2}, \frac{13}{2})$ and $(- \frac{3}{2}, \frac{3}{2})$ .
Key Takeaway: Absolute value in area problems often yields two distinct geometric solutions.

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The Sigma Insight: Area of Triangle

Solution Diagram

Analyzing the Setup

We are given two fixed points,

A (2, 1)

and

B (3, - 2)

. The third vertex

C

lies on the line

y = x + 3

By defining the

x

-coordinate of

C

λ

, we can express the coordinates of

C

(λ, λ + 3)

. This reduces the unknown position of

C

to a single parameter,

λ

The Master Equation

We utilize the coordinate area formula for a triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

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

Given that the area is

5

, we substitute our points

A (2, 1)

B (3, - 2)

, and

C (λ, λ + 3)

into the formula:

\frac{1}{2} ∣2 (- 2 - (λ + 3)) + 3 ((λ + 3) - 1) + λ (1 - (- 2)) ∣ = 5

Simplifying the Expression

Multiplying both sides by

2

, we obtain:

∣2 (- 2 - λ - 3) + 3 (λ + 2) + 3 λ ∣ = 10

Expanding the terms inside the modulus:

∣2 (- λ - 5) + 3 λ + 6 + 3 λ ∣ = 10

∣ - 2 λ - 10 + 3 λ + 6 + 3 λ ∣ = 10

∣4 λ - 4∣ = 10

Solving for Vertex C

The absolute value equation

∣4 λ - 4∣ = 10

yields two distinct cases, representing the two possible positions for vertex

C

Case 1:

4 λ - 4 = 10

4 λ = 14 \Rightarrow λ = \frac{7}{2}

Substituting

λ = \frac{7}{2}

into

y = λ + 3

, we find

y = \frac{7}{2} + 3 = \frac{13}{2}

. Thus, the first possible coordinate is

C_{1} = (\frac{7}{2}, \frac{13}{2})

Case 2:

4 λ - 4 = - 10

4 λ = - 6 \Rightarrow λ = - \frac{3}{2}

Substituting

λ = - \frac{3}{2}

into

y = λ + 3

, we find

y = - \frac{3}{2} + 3 = \frac{3}{2}

. Thus, the second possible coordinate is

C_{2} = (- \frac{3}{2}, \frac{3}{2})

The two possible locations for vertex

C

are $(\frac{7}{2}, \frac{13}{2})$ and $(- \frac{3}{2}, \frac{3}{2})$ .

Common Traps & FAQs

Wait, why do we use the absolute value in the area formula? Isn't area always positive?

You are absolutely right! Area is always positive, which is exactly why we use the absolute value. In coordinate geometry, the expression inside the modulus can result in a negative value depending on the order in which you list the vertices (clockwise vs. counter-clockwise). The absolute value ensures that no matter the orientation, the calculated area remains a positive physical quantity.

Why do we get two answers for C? Does it mean there are two triangles?

Yes, exactly! Think of the line segment AB as the base of your triangle. The area is 5, which means the height of the triangle must be constant. There are two lines parallel to AB at the required distance—one on either side of the line segment. Vertex C can lie on either of these lines, leading to two distinct triangles that both satisfy the area condition.

Can I use the base-height formula instead of the determinant formula?

You certainly can! You would calculate the length of base AB using the distance formula, then find the perpendicular distance from the line

y = x + 3

to the line containing AB. While this is a valid geometric approach, the determinant (or coordinate) formula is generally faster and less prone to algebraic errors in this specific type of problem.

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

Visualized Solution

Visualizing the Given Data

Defining Vertex $C$ Parametrically

The Area Formula

Substituting the Values

Simplifying the Terms

Combining the Terms

Handling the Absolute Value

Solving Case 1

Solving Case 2

Final Conclusion

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Master Equation

Simplifying the Expression

Solving for Vertex C

Similar Questions

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 $A (- 1, 1)$ 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 .

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

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 $P$ .

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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 :

Animated Solution for Mathematics - Straight Lines: .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.

Visualized Solution

Visualizing the Given Data

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

The Area Formula

Substituting the Values

Simplifying the Terms

Combining the Terms

Handling the Absolute Value

Solving Case 1

Solving Case 2

Final Conclusion

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Master Equation

Simplifying the Expression

Solving for Vertex C

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; } 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 sq. units and it lies on the line, 3x+y−4λ=0, then the value of λ is :

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

Defining Vertex $C$ Parametrically

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 .

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 $Δ P A B = 5$ sq. units and it lies on the line, $3 x + y - 4 λ = 0$ , then the value of $λ$ is :