Let for any three distinct consecutive terms a, b, c of an A.P, the lines ax + by + c = 0 be concurr

Q: Why does $a, b, c$ in A.P. force concurrency at $(1, -2)$? I don't see the connection.

Think of the equation $ax + by + c = 0$ as a family of lines. The condition $a - 2b + c = 0$ is exactly what you get if you plug $x=1$ and $y=-2$ into the line equation. It is not a coincidence; it is the definition of the point where all these lines must meet! By comparing the coefficients, you are essentially 'locking' the point that satisfies the equation for any valid $a, b, c$.

Q: Why do we set $\Delta = 0$ for infinite solutions? Can't we just use Gaussian elimination?

You absolutely can use Gaussian elimination! However, in JEE Advanced, time is your currency. Using the determinant condition $\Delta = 0$ is often faster and less prone to algebraic errors when dealing with parameters like $\alpha$ and $\beta$. It is a powerful tool in your arsenal that directly targets the condition for dependency in a system of linear equations.

Q: Do I really need to calculate $\Delta_x, \Delta_y, \Delta_z$?

For infinite solutions, yes, you need the system to be consistent and dependent. If $\Delta = 0$ but the auxiliary determinants are non-zero, the system is inconsistent (no solutions). So, checking $\Delta_x = 0$ is mandatory to find the specific value of $\beta$ that allows for infinite solutions. It is the gatekeeper that ensures the system doesn't just collapse into a contradiction.

Question

Let for any three distinct consecutive terms $a, b, c$ of an A.P, the lines $ax + by + c = 0$ be concurrent at the point $P$ and $Q(\alpha, \beta)$ be a point such that the system of equations $x + y + z = 6, 2x + 5y + \alpha z = \beta$ and $x + 2y + 3z = 4$, has infinitely many solutions. Then $(PQ)^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of A.P.

We start with the lines $ax + by + c = 0$, where $a, b, c$ are in Arithmetic Progression. The beauty of an A.P. is its rigid structure. If $a, b, c$ are in A.P., then the common difference is constant, meaning $b - a = c - b$.

If we rearrange this, we get the elegant constraint:
$$a - 2b + c = 0$$

Now, look at the line equation $ax + by + c = 0$. If we compare this to our constraint $a(1) + b(-2) + c = 0$, the revelation is immediate.

For any values of $a, b, c$ that satisfy the A.P. condition, the equation is satisfied if $x = 1$ and $y = -2$. This means that every single line in this family, regardless of the specific A.P., must pass through the fixed point $P(1, -2)$.

### The Algebra of Infinity

Now, we turn our attention to the system of equations:
$$x + y + z = 6$$
$$2x + 5y + \alpha z = \beta$$
$$x + 2y + 3z = 4$$

We are told this system has infinitely many solutions. In the language of linear algebra, this implies that the planes represented by these equations are not independent; they intersect in a way that creates a line of solutions.

For this to happen, the determinant of the coefficient matrix, $\Delta$, must be zero. Let us construct it:
$$\Delta = \begin{vmatrix} 1 & 1 & 1 \ 2 & 5 & \alpha \ 1 & 2 & 3 \end{vmatrix}$$

Expanding this along the first row, we get $1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5)$. Simplifying this, we find:
$$\Delta = 15 - 2\alpha - 6 + \alpha - 1 = 8 - \alpha$$

Setting $\Delta = 0$, we find $\alpha = 8$. With $\alpha$ in hand, we must find $\beta$. We use the auxiliary determinant $\Delta_x$, replacing the first column with the constants $6, \beta, 4$:
$$\Delta_x = \begin{vmatrix} 6 & 1 & 1 \ \beta & 5 & 8 \ 4 & 2 & 3 \end{vmatrix}$$

Expanding this, we get $6(15 - 16) - 1(3\beta - 32) + 1(2\beta - 20)$. This simplifies to $-6 - 3\beta + 32 + 2\beta - 20 = 6 - \beta$. Setting $\Delta_x = 0$, we find $\beta = 6$. Our point $Q$ is $(8, 6)$.

### The Final Convergence

We have reached the endgame. We have $P(1, -2)$ and $Q(8, 6)$. The problem asks for $(PQ)^2$.

Using the distance formula:
$$(PQ)^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$

Substituting our coordinates, we get $(8 - 1)^2 + (6 - (-2))^2$. This becomes $7^2 + 8^2$, which is $49 + 64$.

The final result is:
$$(PQ)^2 = \mathbf{113}$$

Enter Numerical Value:

Visualized Solution

Condition for A.P.

Finding Point $P$

Condition for Infinite Solutions

Setting up Determinant $Δ$

Solving for $α$

Setting up Determinant $Δ_{x}$

Solving for $β$

Setting up $(P Q)^{2}$

Substituting Coordinates

Final Calculation

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Geometry of A.P

The Algebra of Infinity

The Final Convergence

Similar Questions

If the system of linear equations $2 x + 2 a y + a z = 0 2 x + 3 b y + b z = 0 2 x + 4 cy + cz = 0$ , where $a, b, c \in R$ are non-zero and distinct ; has a non-zero solution, then

If the system of linear equations $2 x + 2 a y + a z = 0, 2 x + 3 b y + b z = 0, 2 x + 4 cy + cz = 0$ , where $a, b, c \in R$ are non-zero and distinct; has non-zero solution, then

If the system of linear equations $7 x + 11 y + a z = 13$ , $5 x + 4 y + 7 z = β$ , $175 x + 194 y + 57 z = 361$ has infinitely many solutions, then $a + β + 2$ is equal to

If the system of equations $(λ - 1) x + (λ - 4) y + λ z = 5$ $λ x + (λ - 1) y + (λ - 4) z = 7$ $(λ + 1) x + (λ + 2) y - (λ + 2) z = 9$ has infinitely many solutions, then $λ^{2} + λ$ is equal to

If the system of equations $2 x + 3 y - z = 5$ , $x + α y + 3 z = - 4$ , $3 x - y + β z = 7$ has infinitely many solutions, then $13 α β$ is equal to

If the system of linear equations $x - 2 y + z = - 4, 2 x + α y + 3 z = 5, 3 x - y + β z = 3$ has infinitely many solutions, then $12 α + 13 β$ is equal to

If the system of equations $2 x + y - z = 5$ , $2 x - 5 y + λ z = μ$ , $x + 2 y - 5 z = 7$ has infinitely many solutions, then $(λ + μ)^{2} + (λ - μ)^{2}$ is equal to

If the system of equation $2 x + λ y + 3 z = 5$ , $3 x + 2 y - z = 7$ , $4 x + 5 y + μ z = 9$ has infinitely many solutions, then $(λ^{2} + μ^{2})$ is equal to :

If the system of linear equations $x + 2 a y + a z = 0$ ; $x + 3 b y + b z = 0$ ; $x + 4 cy + cz = 0$ ; has a non-zero solution, then $a, b, c$

For a real number $a$ , if the system $1 a a^{2} a 1 a a^{2} a 1 x y z = 1 - 1 1$ of linear equations, has infinitely many solutions, then $1 + a + a^{2} =$

Enter Numerical Value:

Visualized Solution

Condition for A.P.

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

Condition for Infinite Solutions

Setting up Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

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

Setting up Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx​

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

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (PQ)2

Substituting Coordinates

Final Calculation

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Geometry of A.P

The Algebra of Infinity

The Final Convergence

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