A rod of length eight units moves such that its ends A and B always lie on the lines x-y+2=0 and y+2

Q: Why did we choose 'a' and 'b' as parameters? Can't we use the angle of the rod?

You absolutely could use the angle $\theta$! However, in coordinate geometry, parameterizing based on the lines themselves is often more direct. It avoids trigonometric identities and keeps the algebra strictly polynomial, which is usually safer for locus problems.

Q: I got lost in the algebra during the substitution. Is there a trick?

Algebra is the language of physics. The trick is to isolate your variables $a$ and $b$ *before* plugging them into the distance formula. Don't try to substitute everything at once. Break it down into smaller, manageable chunks.

Q: Why do we need to eliminate $a$ and $b$?

The locus is the set of all points $(x, y)$ that satisfy the condition. If we leave $a$ and $b$ in the equation, we are describing the rod's position, not the path of point $P$. We must eliminate the 'dummy' variables to find the relationship between $x$ and $y$ alone.

Question

A rod of length eight units moves such that its ends A and B always lie on the lines $x-y+2=0$ and $y+2=0$ respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2:1 is $9(x^{2}+\alpha y^{2}+\beta xy+\gamma x+28y)-76=0$, then $\alpha-\beta-\gamma$ is equal to :

Options:

(A) 22

(B) 21

(C) 23

(D) 24

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving an equation; we are observing a dance. Imagine a rigid rod, a physical object of length $8$, trapped between two lines.

It cannot escape, it cannot stretch, and it cannot shrink. It is a prisoner of geometry. Our task is to trace the path of a specific point $P$ on this rod, which is the essence of a locus problem.

### Defining the Players

Before we can analyze the motion, we must name the actors. We have two lines: $L_1: x - y + 2 = 0$ and $L_2: y + 2 = 0$.

The rod has two ends, $A$ and $B$. Point $A$ is constrained to $L_1$, and point $B$ is constrained to $L_2$.

Let the $x$-coordinate of $A$ be $a$. Since $A$ lies on $x - y + 2 = 0$, we find $y = x + 2$, so $A = (a, a + 2)$.

Similarly, $B$ lies on $y + 2 = 0$, meaning its $y$-coordinate is fixed at $-2$. Let its $x$-coordinate be $b$. Thus, $B = (b, -2)$.

### The Rigid Constraint

Nature imposes laws, and in this problem, the law is the rod's length. The distance $AB$ is constant at $8$, so $AB^2 = 64$.

Using the distance formula, we write:
$$(a - b)^2 + (a + 2 - (-2))^2 = 64$$

Simplifying this, we get:
$$(a - b)^2 + (a + 4)^2 = 64$$

This is our **golden constraint**. It is the anchor that prevents the rod from flying off into infinity.

### The Bridge (Section Formula)

Now, we introduce point $P(x, y)$, which divides $AB$ in a $2:1$ ratio. The section formula is our bridge between the rod's endpoints and the locus we seek.

For the $x$-coordinate:
$$x = \frac{1 \cdot a + 2 \cdot b}{2 + 1} = \frac{a + 2b}{3}$$

For the $y$-coordinate:
$$y = \frac{1 \cdot (a + 2) + 2 \cdot (-2)}{2 + 1} = \frac{a - 2}{3}$$

### The Algebraic Dance

We must now eliminate $a$ and $b$. From the $y$ equation, we find $a = 3y + 2$.

Substituting this into the $x$ equation, we solve for $b$:
$$3x = a + 2b \Rightarrow 2b = 3x - a = 3x - (3y + 2) = 3x - 3y - 2$$

Thus, $b = \frac{3x - 3y - 2}{2}$. Now, we return to our golden constraint: $(a - b)^2 + (a + 4)^2 = 64$.

After careful substitution and expansion, we arrive at the equation:
$$9(3y - x + 2)^2 + 36(y + 2)^2 = 256$$

### The Grand Finale

Expanding this yields:
$$9x^2 + 117y^2 - 54xy - 36x + 252y - 76 = 0$$

Factoring out the $9$, we match the form $9(x^2 + \alpha y^2 + \beta xy + \gamma x + 28y) - 76 = 0$. By comparison, we find $\alpha = 13$, $\beta = -6$, and $\gamma = -4$.

The final calculation, $\alpha - \beta - \gamma = 13 - (-6) - (-4)$, gives us **23**. You have successfully navigated the constraints and distilled the geometry into a single, elegant number.

Select Answer:

Visualized Solution

Visualizing the Setup

Parametrizing Point $A$

Parametrizing Point $B$

The Distance Constraint

Applying Section Formula

Solving for Variable $a$

Solving for Variable $b$

Calculating $(a - b)$

Calculating $(a + 4)$

Substitution into Distance Equation

Final Algebraic Simplification

Comparing and Final Answer

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

Defining the Players

The Rigid Constraint

The Bridge (Section Formula)

The Algebraic Dance

The Grand Finale

Similar Questions

A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

If the equation of the locus of a point equidistant from the point $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then the value of 'c' is

If the line $2 x + y = k$ passes through the point which divides the line segment joining the points $(1, 1)$ and $(2, 4)$ in the ratio $3 : 2$ , then $k$ equals

Let $A (- 2, - 1), B (1, 0), C (α, β)$ and $D (γ, δ)$ be the vertices of a parallelogram $A B C D$ . If the point $C$ lies on $2 x - y = 5$ and the point $D$ lies on $3 x - 2 y = 6$ , then the value of $∣ α + β + γ + δ ∣$ is equal to _______.

For $a > b > c > 0$ , the distance between $(1, 1)$ and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less than $22$ . Then

Let $S$ be a square of unit area. Consider any quadrilateral which has one vertex on each side of $S$ . If $a, b, c$ , and $d$ denote the lengths of the sides of the quadrilateral, prove that $2 \leq a^{2} + b^{2} + c^{2} + d^{2} \leq 4$ .

If $(1, 2), (4, 6), (5, 7)$ and $(a, b)$ are the vertices of a parallelogram $P QR S$ , then

A triangle with vertices $(4, 0), (- 1, - 1), (3, 5)$ is