Let the line x + y = 1$x + y = 1$ meet the axes of x$x$ and y$y$ 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 (4)/(9)$\frac{4}{9}$ of the area of the triangle OAB and AN: NB = \lambda : 1, then the sum of all possible value(s) of λ$\lambda$ is :

Related Formula

Where a$a$ is the side length.

Core Logic

Let the side of the equilateral triangle be a$a$. Let θ$\theta$ be the angle between the side AC$AC$ and the parallel line L₂$L_2$. Then, the angle between the side AB$AB$ and the parallel line L₁$L_1$ can be expressed via alternate geometry. Given distances from A$A$ to the lines form right-angled triangles.

Step 1: Set up geometric projections

From vertex C$C$ to line passing through A$A$ parallel to L₁, L₂$L_1, L_2$, the perpendicular distance is 3$3$. In the right triangle formed, we have:

Similarly, point B$B$ lies on L₁$L_1$. The perpendicular distance from A$A$ to L₁$L_1$ is 6$6$. However, combining the overall heights between the parallel lines, the total distance between L₁$L_1$ and L₂$L_2$ is 6 + 3 = 9$6 + 3 = 9$. The projection of side BC$BC$ (which connects L₁$L_1$ and L₂$L_2$) vertically is 9$9$.

Step 2: Solve the trigonometric system

Expand (60° + θ)$\sin(60^{\circ} + \theta)$:

Substitute θ = (3)/(a)$\sin \theta = \frac{3}{a}$ and θ = √(1 - ² θ) = √(1 - (9)/(a²))$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{9}{a^2}}$:

Squaring both sides:

Step 3: Calculate Area

Pattern Recognition

When a rigid polygon (like an equilateral triangle or square) is wedged between parallel lines, set a base orientation angle θ$\theta$ for one edge and use rotational shifts (e.g., 60° + θ$60^{\circ} + \theta$) to project heights. Expanding the sine addition formula instantly yields the side length.

Chapter Mix

Class 11 Maths: Straight Lines Class 11 Maths: Trigonometric Functions

Related Formula

Orthocenter property: Altitude from A$A$ is perpendicular to BC$BC$, Altitude from B$B$ is perpendicular to AC$AC$. Concurrency of lines: ax+by+c=0$ax+by+c=0$ passes through (x₀, y₀)$(x_0, y_0)$ if ax₀+by₀+c=0$ax_0+by_0+c=0$ holds.

Core Logic

Statement 1: Let third vertex be C(h,k)$C(h,k)$ and orthocenter O(0,0)$O(0,0)$.

• Slope AO ·$AO \cdot$ Slope BC = -1 ((-1)/(5)) · ((k-3)/(h+2)) = -1 5h - k + 13 = 0$BC = -1 \implies \left(\frac{-1}{5}\right) \cdot \left(\frac{k-3}{h+2}\right) = -1 \implies 5h - k + 13 = 0$
• Slope BO ·$BO \cdot$ Slope AC = -1 ((3)/(-2)) · ((k+1)/(h-5)) = -1 4k = 7h$AC = -1 \implies \left(\frac{3}{-2}\right) \cdot \left(\frac{k+1}{h-5}\right) = -1 \implies 4k = 7h$

Solving simultaneously gives h = -4, k = -7 C(-4,-7)$h = -4, k = -7 \implies C(-4,-7)$. Statement 1 is correct.

Statement 2: 2a, b, c$2a, b, c$ in A.P. 2b = 2a + c 2a - 2b + c = 0$\implies 2b = 2a + c \implies 2a - 2b + c = 0$. Comparing with ax + by + c = 0$ax + by + c = 0$ gives x = 2, y = -2$x = 2, y = -2$. Statement 2 is correct.

Step 1: Final Conclusion

Both (S1) and (S2) are correct statements.

Pattern Recognition

Orthocenter coordinates yield perpendicularity conditions via slopes. Concurrency follows directly from linear relation between line coefficients.

Chapter Mix

Class 11 Maths: Straight Lines

Related Formula

Core Logic

Any line that divides the area of a rectangle into two equal parts must pass through the center of the rectangle.

Step 1: Equation of Line L

The line L$L$ is perpendicular to 3x + y + 6 = 0$3x + y + 6 = 0$. The slope of this reference line is -3$-3$. Thus, the slope of L$L$ is m = (1)/(3)$m = \frac{1}{3}$. The equation of L$L$ passing through ((3)/(2), 2)$\left(\frac{3}{2}, 2\right)$ is:

Multiplying by 6 gives:

Step 2: Calculate Distance

We need the perpendicular distance from ((1)/(2), -5)$\left(\frac{1}{2}, -5\right)$ to the line 2x - 6y + 9 = 0$2x - 6y + 9 = 0$:

Pattern Recognition

Area bisectors for symmetric geometric figures (rectangles, circles, ellipses) always pass exactly through their geometric center.

Chapter Mix

Class 11 Maths: Straight Lines

Related Formula

In a rhombus, diagonals bisect each other. Therefore, the midpoint of diagonal AC$AC$ is the same as the midpoint of diagonal BD$BD$.

Core Logic

So, the midpoint is (-1, -2)$(-1, -2)$.

Let the coordinates of B$B$ and D$D$ be (α, β)$(\alpha, \beta)$ and (γ, δ)$(\gamma, \delta)$ respectively. Since the midpoint of BD$BD$ is also (-1, -2)$(-1, -2)$:

Step 1: Final Calculation

We need the absolute value of the sum of the coordinates of B$B$ and D$D$:

Pattern Recognition

The extra information regarding the parallel line 7x - y = 14$7x - y = 14$ is completely redundant for finding the sum of the coordinates. Always check if a basic geometric property (like diagonals bisecting) circumvents heavy calculations.

Chapter Mix

Class 11 Maths: Straight Lines

Related Formula

Angle bisector divides the opposite side in the ratio of adjacent sides: (BD)/(DC) = (AB)/(BC)$\frac{BD}{DC} = \frac{AB}{BC}$ (Internal Bisector Theorem).

Core Logic

Step 1: Side Lengths

Ratio (AB)/(BC) = √(2) 5√(2)3 = (3)/(5)$\frac{AB}{BC} = \frac{\sqrt{2}}{\frac{5\sqrt{2}}{3}} = \frac{3}{5}$.

Step 2: Coordinates of D

Point D$D$ divides AC$AC$ internally in ratio 3:5$3:5$.

D = ((3)/(2), (1)/(2))$D = \left(\frac{3}{2}, \frac{1}{2}\right)$

Step 3: Equation of Bisector

The bisector passes through B(2,-1)$B(2,-1)$ and D(3/2, 1/2)$D(3/2, 1/2)$. Slope m = (1/2 - (-1))/(3/2 - 2) = (3/2)/(-1/2) = -3$m = \frac{1/2 - (-1)}{3/2 - 2} = \frac{3/2}{-1/2} = -3$ Equation: y - (-1) = -3(x - 2)$y - (-1) = -3(x - 2)$ y + 1 = -3x + 6 ⇒ 3x + y = 5$y + 1 = -3x + 6 \Rightarrow 3x + y = 5$

Step 4: Final Value Calculation

Comparing 3x + y = 5$3x + y = 5$ with α x + β y = 5$\alpha x + \beta y = 5$, we get α = 3$\alpha = 3$, β = 1$\beta = 1$.

Pattern Recognition

When asked for a specific angle bisector in a coordinate triangle, using the Angle Bisector Theorem (ratio trick) to find a second point is much faster than computing formulas of bisectors between lines.

Chapter Mix

Class 11 Maths: Straight Lines