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 :

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Core Logic

The initial line makes an angle of 30°$30^{\circ}$ with the positive x-axis. It is rotated clockwise by 15°$15^{\circ}$ about the point A(9, 0)$A(9, 0)$. The new angle made by the line with the positive direction of the x-axis is 30° - 15° = 15°$30^{\circ} - 15^{\circ} = 15^{\circ}$.

Step 1: Equation of the new line

The equation of the line passing through A(9, 0)$A(9, 0)$ with a slope of 15°$\tan 15^{\circ}$ is:

We know that 15° = 2 - √(3)$\tan 15^{\circ} = 2 - \sqrt{3}$.

Step 2: Rearranging to match options

Dividing by (2 - √(3))$(2 - \sqrt{3})$:

Notice that 2 - √(3) = -(√(3) - 2)$2 - \sqrt{3} = -(\sqrt{3} - 2)$. Thus:

Pattern Recognition

A clockwise rotation decreases the angle of inclination. Calculate the new angle, find its tangent, and carefully algebraicize the denominator to match the given option forms.

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Class 11 Maths: Straight Lines

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Distance in parametric form: x = x₁ + r θ, y = y₁ + r θ$x = x_1 + r\cos\theta, y = y_1 + r\sin\theta$

Core Logic

Let Orthocentre C(-6, -8)$C(-6, -8)$ and Circumcentre B(3, 4)$B(3, 4)$. Centroid A(a, b)$A(a, b)$ divides CB$CB$ in 2:1$2:1$.

So, P(2a+3, 7b+5) = (3, 5)$P(2a+3, 7b+5) = (3, 5)$.

The line along which distance is measured is parallel to x - 2y - 1 = 0$x - 2y - 1 = 0$, giving slope m = θ = (1)/(2)$m = \tan\theta = \frac{1}{2}$. Using parametric coordinates from P(3,5)$P(3,5)$:

Substitute into the target line 2x + 3y - 4 = 0$2x + 3y - 4 = 0$:

From θ = 1/2$\tan\theta = 1/2$, we get θ = 1√(5)$\sin\theta = \frac{1}{\sqrt{5}}$ and θ = 2√(5)$\cos\theta = \frac{2}{\sqrt{5}}$.

Pattern Recognition

Standard Euler line property: O, G, C$O, G, C$ are collinear and G$G$ divides OC$OC$ in 2:1$2:1$. Use parametric equation to find intersection distance directly without finding the intersection point.

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Class 11 Maths: Straight Lines

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Core Logic

Given diagonals AC$AC$ and BD$BD$ bisect each other:

Point C(α, β)$C(\alpha, \beta)$ lies on 2x - y = 5 2α - β = 5 (3)$2x - y = 5 \implies 2\alpha - \beta = 5 \quad \dots (3)$ Point D(γ, δ)$D(\gamma, \delta)$ lies on 3x - 2y = 6 3γ - 2δ = 6 (4)$3x - 2y = 6 \implies 3\gamma - 2\delta = 6 \quad \dots (4)$

From (1) and (2), substitute γ = α - 3$\gamma = \alpha - 3$ and δ = β - 1$\delta = \beta - 1$ into (4):

Solve (3) and (5): From (3), β = 2α - 5$\beta = 2\alpha - 5$. Substitute in (5):

So, β = 2(-3) - 5 = -11$\beta = 2(-3) - 5 = -11$. From earlier substitutions:

Sum of variables:

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Class 11 Maths: Straight Lines

Core Logic

Let E$E$ be the midpoint of the diagonals AC$AC$ and BD$BD$. Since ABCD$ABCD$ is a parallelogram, the diagonals bisect each other.

Step 1: Apply Midpoint on AC

Midpoint E$E$ from AC$AC$: ( (α+γ)/(2), (β+δ)/(2) )$\left( \frac{\alpha+\gamma}{2}, \frac{\beta+\delta}{2} \right)$. Equating both:

Step 2: Final Value

The expression requires 2(α + β + γ + δ)$2(\alpha + \beta + \gamma + \delta)$.

(Note: Additional conditions like AB = √(10)$AB = \sqrt{10}$ and the line equation are extraneous data not needed to find the sum).

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Class 11 Maths: Straight Lines