Let the lines 3x-4y-α=0, 8x-11y-33=0 and 2x-3y+λ=0 be concurrent. If the image of the point (1,2) in

Q: Wait, why does the midpoint of the point and its image have to lie on the mirror line?

Think of the mirror line as the axis of symmetry. By definition, the reflection of a point $P$ across a line $L$ is a point $Q$ such that the line $L$ is the perpendicular bisector of the segment $PQ$. Since $L$ bisects $PQ$, the point where they intersect—the midpoint—must lie on the line $L$.

Q: Why can't we just use the distance formula to find the intersection point?

The distance formula is great for finding lengths, but it doesn't directly give you the intersection point of two lines. For concurrency, we need the specific coordinates $(x, y)$ where the lines meet. Solving the system of linear equations is the most direct and robust algebraic method to find that unique point.

Q: Is there a specific order I should follow to solve this?

Yes! The problem gives you the image $Q$ to help you find the unknown $\lambda$ in the third line. Once you have the complete equation for $L_3$, you have two lines ($L_2$ and $L_3$) with known coefficients. Solving those gives you the intersection point, which then unlocks the final variable $\alpha$ in $L_1$. Always look for the 'chain' of information!

Question

Let the lines $3x-4y-\alpha=0,$ $8x-11y-33=0$ and $2x-3y+\lambda=0$ be concurrent. If the image of the point $(1,2)$ in the line $2x-3y+\lambda=0$ is $(\frac{57}{13},\frac{-40}{13})$ then $\alpha$ is equal to

Options:

(A) 84

(B) 113

(C) 91

(D) 101

Sigma Percentile · Accepted Answer

### The Geometry of Reflections and Concurrency

Welcome, fellow explorer of the coordinate plane! Today, we are going to unravel a beautiful problem that weaves together two fundamental concepts: the symmetry of reflections and the elegance of concurrent lines.

Imagine you are standing on a vast, infinite grid. You have a point $P(1, 2)$ and a mirror line $L_3: 2x - 3y + \lambda = 0$. When you look into this mirror, you see an image $Q(\frac{57}{13}, -\frac{40}{13})$.

Our mission is to find the hidden parameter $\alpha$ in the first line $L_1: 11x - 6y - \alpha = 0$. Let's embark on this journey step by step.

### Phase 1: The Mirror Mystery

First, we must decode the mirror line $L_3$. We know that when a point $P$ is reflected across a line to form an image $Q$, the line itself acts as the perpendicular bisector of the segment $PQ$.

This is a powerful geometric truth! It implies that the midpoint $M$ of the segment connecting $P$ and $Q$ must lie exactly on the mirror line $L_3$.

Let's calculate the coordinates of this midpoint $M$ using the midpoint formula, $M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$. Plugging in our values for $P(1, 2)$ and $Q(\frac{57}{13}, -\frac{40}{13})$:

$$x_M = \frac{1 + \frac{57}{13}}{2} = \frac{\frac{13+57}{13}}{2} = \frac{70}{26} = \frac{35}{13}$$

$$y_M = \frac{2 + (-\frac{40}{13})}{2} = \frac{\frac{26-40}{13}}{2} = -\frac{14}{26} = -\frac{7}{13}$$

Now that we have $M(\frac{35}{13}, -\frac{7}{13})$, we know it must satisfy the equation of the mirror line $L_3: 2x - 3y + \lambda = 0$. Substituting these coordinates in, we get:

$$2\left(\frac{35}{13}\right) - 3\left(-\frac{7}{13}\right) + \lambda = 0$$

$$\frac{70}{13} + \frac{21}{13} + \lambda = 0$$

$$\frac{91}{13} + \lambda = 0$$

$$7 + \lambda = 0 \Rightarrow \lambda = -7$$

We have successfully unlocked the identity of our third line: $L_3: 2x - 3y - 7 = 0$. The mirror is now fully defined!

### Phase 2: The Concurrency Connection

Now, let's tackle the concept of concurrency. When we say three lines are concurrent, we mean they all intersect at one single, common point $C$.

We already have the equations for $L_2: 8x - 11y - 33 = 0$ and our newly found $L_3: 2x - 3y - 7 = 0$. To find the intersection point $C$, we simply solve these two equations as a system:

1. $8x - 11y = 33$
2. $2x - 3y = 7$

To eliminate $x$, let's multiply the second equation by $4$, giving us $8x - 12y = 28$. Now, subtract this from the first equation:

$$(8x - 11y) - (8x - 12y) = 33 - 28$$

$$y = 5$$

With $y=5$ in hand, we substitute it back into $2x - 3y = 7$ to find $x$:

$$2x - 3(5) = 7$$

$$2x - 15 = 7$$

$$2x = 22 \Rightarrow x = 11$$

Our intersection point $C$ is $(11, 5)$. This is the golden key that unlocks the final part of our puzzle.

### Phase 3: The Final Reveal

We are told that all three lines are concurrent, which means the point $C(11, 5)$ must also lie on the first line $L_1: 11x - 6y - \alpha = 0$. This is the moment of truth.

We substitute $x=11$ and $y=5$ into the equation for $L_1$:

$$11(11) - 6(5) - \alpha = 0$$

$$121 - 30 - \alpha = 0$$

$$91 - \alpha = 0$$

$$\alpha = 91$$

And there it is! Through the symmetry of reflection and the intersection of lines, we have arrived at **$\alpha = 91$**.

It is a beautiful result, isn't it? Remember, in coordinate geometry, every equation is a story, and every intersection is a meeting of paths. Keep practicing, keep visualizing, and you will master these concepts with ease!

Animated Solution for Mathematics - Straight Lines: Let the lines $3 x - 4 y - α = 0,$ $8 x - 11 y - 33 = 0$ and $2 x - 3 y + λ = 0$ be concurrent. If the image of the point $(1, 2)$ in the line $2 x - 3 y + λ = 0$ is $(\frac{57}{13}, \frac{- 40}{13})$ then $α$ is equal to

Select Answer:

Visualized Solution

Visualizing the Problem

The Midpoint Property

Calculating Midpoint $x$ -coordinate

Calculating Midpoint $y$ -coordinate

Finding $λ$

Solving for $λ$

The Concept of Concurrency

Solving the System of Equations

Finding the $y$ -coordinate

Finding the $x$ -coordinate

Finding $α$

Final Calculation

Summary and Key Takeaway

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of Reflections and Concurrency

Phase 1

Phase 2

Phase 3

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