Let the line x+y=1 meet the axes of x and y at A and B, respectively. A right angled triangle AMN is

Q: Wait, how do we know the right angle is at $N$?

Great observation! In many JEE problems, the geometry is constrained by the points' locations. Since $M$ is on the y-axis and $N$ is on the hypotenuse $AB$, if the right angle were at $A$, $MN$ would have to be perpendicular to $AB$, which is the same condition. If it were at $M$, $AM$ would be perpendicular to $MN$. The problem implies the configuration where $N$ is the vertex of the right angle. Always sketch the diagram first to see which vertex makes sense!

Q: Why do we use the slope of $AB$ as $-1$?

Look at the line equation $x+y=1$. Rewrite it as $y = -x + 1$. This is in the slope-intercept form $y = mx + c$, where the slope is clearly $-1$. Since $MN \perp AB$, the product of their slopes must be $-1$. Thus, the slope of $MN$ is $1$. It's a fundamental property of perpendicular lines!

Q: Is there a faster way than the section formula?

You could use similar triangles or ratios of areas, but the section formula is the most robust method for JEE. It connects the coordinate geometry directly to the ratio $\lambda$. Don't look for shortcuts; master the standard tools first, and you will find the path to the answer becomes much clearer.

Question

Let the line $x+y=1$ meet the axes of x and y at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is $\frac{4}{9}$ of the area of the triangle OAB and AN : NB = $\lambda$ : 1, then the sum of all possible value(s) of $\lambda$ is :

Options:

(A) 2

(B) $\frac{5}{2}$

(C) $\frac{1}{2}$

(D) $\frac{13}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dissect a beautiful problem that sits at the intersection of coordinate geometry and pure intuition.

We are given a line, $x+y=1$, which acts as our canvas. It carves out a right-angled triangle $	riangle OAB$ with the origin $O(0,0)$, the x-axis intercept $A(1,0)$, and the y-axis intercept $B(0,1)$.

The area of this triangle is trivial to calculate:

$$Area = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$$

Keep this value in your mind; it is the benchmark for everything that follows.

### The Constraint: The Right Angle at $N$

Now, we introduce a new player: $	riangle AMN$. We are told it is a right-angled triangle inscribed within $	riangle OAB$.

Point $M$ is constrained to the y-axis, meaning its coordinates are $(0, m)$. Point $N$ is constrained to the hypotenuse $AB$, so its coordinates must satisfy $x+y=1$, which we can write as $(x, 1-x)$.

The problem tells us $	riangle AMN$ is right-angled. By analyzing the geometry, we identify that the right angle must be at $N$.

If $MN \perp AB$, then the slope of $MN$ must be the negative reciprocal of the slope of $AB$. Since the slope of $AB$ is $-1$, the slope of $MN$ must be $1$. This is the algebraic key that unlocks the door.

### The Algebraic Dance

Let us calculate the slope of $MN$ using the coordinates $M(0, m)$ and $N(x, 1-x)$. The slope is:

$$\frac{(1-x) - m}{x - 0} = 1$$

Solving this simple linear equation, we get $1 - x - m = x$, which simplifies to $2x = 1 - m$, or $x = \frac{1-m}{2}$.

Consequently, the y-coordinate of $N$ is:

$$y = 1 - x = 1 - \frac{1-m}{2} = \frac{1+m}{2}$$

We have now successfully parameterized the position of $N$ in terms of $m$. This is the power of coordinate geometry—turning a visual problem into a solvable algebraic one.

### The Area Calculation

We are told the area of $	riangle AMN$ is $\frac{4}{9}$ of the area of $	riangle OAB$. Since the area of $	riangle OAB$ is $\frac{1}{2}$, the target area for $	riangle AMN$ is:

$$\frac{4}{9} 	imes \frac{1}{2} = \frac{2}{9}$$

Using the distance formula, we find the lengths of the legs $AN$ and $MN$:

$$AN = \sqrt{\left(1 - \frac{1-m}{2}ight)^2 + \left(0 - \frac{1+m}{2}ight)^2} = \frac{1+m}{\sqrt{2}}$$

$$MN = \sqrt{\left(\frac{1-m}{2} - 0ight)^2 + \left(\frac{1+m}{2} - might)^2} = \frac{1-m}{\sqrt{2}}$$

The area of $	riangle AMN$ is:

$$\frac{1}{2} 	imes AN 	imes MN = \frac{1}{2} 	imes \frac{1+m}{\sqrt{2}} 	imes \frac{1-m}{\sqrt{2}} = \frac{1-m^2}{4}$$

Equating this to $\frac{2}{9}$, we get $\frac{1-m^2}{4} = \frac{2}{9}$, which leads to $9(1-m^2) = 8$, or $9m^2 = 1$. Thus, $m = \frac{1}{3}$.

### The Final Ratio

With $m = \frac{1}{3}$, we find the coordinates of $N$:

$$x = \frac{1 - 1/3}{2} = \frac{1}{3}, \quad y = \frac{1 + 1/3}{2} = \frac{2}{3}$$

So, $N$ is at $(\frac{1}{3}, \frac{2}{3})$. The problem states that $N$ divides $AB$ in the ratio $\lambda : 1$.

Using the section formula, the x-coordinate of $N$ is:

$$\frac{1(1) + \lambda(0)}{\lambda+1} = \frac{1}{\lambda+1}$$

Setting this equal to $\frac{1}{3}$, we find $\lambda + 1 = 3$, which gives $\lambda = 2$.

The sum of all possible values is simply **2**. You have navigated the geometry, mastered the algebra, and arrived at the truth.

Select Answer:

Visualized Solution

Visualizing Triangle $△ O A B$

Calculating Area of $△ O A B$

Target Area of $△ A M N$

Defining Positions of $M$ and $N$

Slope Relationship for $M N ⊥ A B$

Finding Coordinates of $N$

Calculating Lengths $A N$ and $M N$

Solving for $m$ using Area

Finding Final Coordinates of $N$

Calculating the Ratio $λ$

Conclusion and Final Sum

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Constraint

The Algebraic Dance

The Area Calculation

The Final Ratio

Similar Questions

Let $A B C$ be an isosceles triangle in which $A$ is at $(- 1, 0)$ , $∠ A = \frac{2 π}{3}$ , $A B = A C$ and $B$ is on the positive $x$ -axis. If $B C = 43$ and the line $B C$ intersects the line $y = x + 3$ at $(α, β)$ , then $\frac{β ^{4}}{α ^{2}}$ is :

Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2 x = 2$ such that $Δ A B C$ is an equilateral triangle. Then, the area of the $Δ A B C$ is

Let the area of the triangle formed by a straight Line $L : x + b y + c = 0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line $L$ makes an angle of $4 5^{\circ}$ with the positive $x$ -axis, then the value of $b^{2} + c^{2}$ is:

Let a point A lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $A B C$ , where the points $B$ and $C$ lie on the lines $L_{1}$ and $L_{2}$ , respectively, is :

A line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $O P Q$ , where $O$ is the origin. If the area of the triangle $O P Q$ is least, then the slope of the line $P Q$ is

Let $(h, k)$ be a fixed point, where $h > 0, k > 0$ . A straight line passing through this point cuts the positive direction of the coordinate axes at the points $P$ and $Q$ . Find the minimum area of the triangle $O P Q, O$ being the origin.

Determine all values of $α$ for which the point $(α, α^{2})$ lies inside the triangle formed by the lines $2 x + 3 y - 1 = 0$ , $x + 2 y - 3 = 0$ , $5 x - 6 y - 1 = 0$ .

The number of integral values of m so that the abscissa of point of intersection of lines $3 x + 4 y = 9$ and $y = m x + 1$ is also an integer, is :

Select Answer:

Visualized Solution

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Calculating Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Target Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △AMN

Defining Positions of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Slope Relationship for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MN⊥AB

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

Calculating Lengths .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AN and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MN

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m using Area

Finding Final Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

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