Key Concepts for Lines Concurrent at a Point

When a family of lines $ax+by+c=0$ passes through a fixed point, it means that there exists a specific point $(x_0, y_0)$ such that $ax_0+by_0+c=0$ for all permissible values of $a, b, c$. Often, the relationship between $a, b, c$ provides the clue to find this fixed point.

Key Concepts for System of Linear Equations with Infinitely Many Solutions

For a system of three linear equations in three variables:

$$\begin{aligned} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{aligned}$$

Let $D$ be the determinant of the coefficient matrix:

$$D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$$

And let $D_x, D_y, D_z$ be the determinants formed by replacing the respective coefficient columns with the constant terms column. For the system to have infinitely many solutions, the necessary and sufficient conditions are: $D = 0 \quad \text{and} \quad D_x = 0 \quad \text{and} \quad D_y = 0 \quad \text{and} \quad D_z = 0$ If $D=0$ and at least one of $D_x, D_y, D_z$ is non-zero, then the system has no solution.

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Step 1: Finding the Point of Concurrency P

We are given that $a, b, c$ are three distinct consecutive terms of an Arithmetic Progression (A.P.). Explanation: In an A.P., the middle term is the average of its neighbors. This relationship gives us a constraint on $a, b, c$. $2b = a+c$ Rearranging this equation, we get: $a - 2b + c = 0$ Now, consider the equation of the line $ax+by+c=0$. Explanation: We need to find a point $(x, y)$ that satisfies this equation regardless of the specific values of $a, b, c$, as long as they satisfy the A.P. condition. We can compare the A.P. condition with the line equation. If we set $x=1$ and $y=-2$ in the line equation, we get: $a(1) + b(-2) + c = 0$ $a - 2b + c = 0$ This is exactly the condition we derived for $a, b, c$ being in an A.P. Explanation: Since the point $(1, -2)$ satisfies the line equation $ax+by+c=0$ whenever $a, b, c$ are in A.P., it means all such lines pass through this fixed point. Therefore, the point of concurrency $P$ is $(1, -2)$.

Tip: This is a common pattern. If you have a linear relation between coefficients like $Aa+Bb+Cc=0$ and a line $ax+by+c=0$, the fixed point is $(A, B)$ if the relation is $Aa+Bb+c=0$, or it can be derived by matching coefficients. Here, $a-2b+c=0$ can be seen as $a(1)+b(-2)+c(1)=0$. If we consider $ax+by+cz=0$ (a plane equation) and $x=1, y=-2, z=1$, this would be satisfied. For a line $ax+by+c=0$, we directly compare $a(1)+b(-2)+c(1)=0$ with $ax_0+by_0+c=0$.

So, we have $P = (1, -2)$.

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Step 2: Finding the Point Q$(\alpha, \beta)$

We are given a system of linear equations:

1. $x+y+z=6$
2. $2x+5y+\alpha z=\beta$
3. $x+2y+3z=4$

This system has infinitely many solutions. Explanation: As per the key concept, for infinitely many solutions, all determinants $D, D_x, D_y, D_z$ must be zero. We will use this to find $\alpha$ and $\beta$.

First, let's calculate the determinant of the coefficient matrix, $D$:

$$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix}$$

Explanation: We set $D=0$ to find $\alpha$. Expanding the determinant along the first row:

$$D = 1 \cdot \begin{vmatrix} 5 & \alpha \\ 2 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & \alpha \\ 1 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 5 \\ 1 & 2 \end{vmatrix}$$ $$D = 1(5 \cdot 3 - \alpha \cdot 2) - 1(2 \cdot 3 - \alpha \cdot 1) + 1(2 \cdot 2 - 5 \cdot 1)$$ $$D = (15 - 2\alpha) - (6 - \alpha) + (4 - 5)$$ $$D = 15 - 2\alpha - 6 + \alpha - 1$$ $$D = 8 - \alpha$$

For infinitely many solutions, $D=0$: $8 - \alpha = 0 \implies \alpha = 8$

Now, we need to find $\beta$. We use the condition that one of the other determinants ($D_x, D_y,$ or $D_z$) must also be zero. Let's calculate $D_x$ (which is $D_1$ in the original solution's notation, replacing the first column with the constant terms).

$$D_x = \begin{vmatrix} 6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3 \end{vmatrix}$$

Explanation: Substitute the value of $\alpha=8$ we just found into $D_x$.

$$D_x = \begin{vmatrix} 6 & 1 & 1 \\ \beta & 5 & 8 \\ 4 & 2 & 3 \end{vmatrix}$$

Now, expand this determinant along the first row:

$$D_x = 6 \cdot \begin{vmatrix} 5 & 8 \\ 2 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} \beta & 8 \\ 4 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} \beta & 5 \\ 4 & 2 \end{vmatrix}$$ $$D_x = 6(5 \cdot 3 - 8 \cdot 2) - 1(\beta \cdot 3 - 8 \cdot 4) + 1(\beta \cdot 2 - 5 \cdot 4)$$ $$D_x = 6(15 - 16) - 1(3\beta - 32) + 1(2\beta - 20)$$ $$D_x = 6(-1) - 3\beta + 32 + 2\beta - 20$$ $$D_x = -6 - 3\beta + 32 + 2\beta - 20$$ $$D_x = 6 - \beta$$

For infinitely many solutions, $D_x=0$: $6 - \beta = 0 \implies \beta = 6$

So, the point $Q(\alpha, \beta)$ is $(8, 6)$.

Common Mistake: Forgetting to substitute the value of $\alpha$ (or any other found variable) into subsequent determinant calculations can lead to incorrect values. Also, be careful with signs during determinant expansion.

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**Step 3: Calculating the Squared Distance $(PQ)^2$}

We have the coordinates of point $P=(1, -2)$ and point $Q=(8, 6)$. Explanation: We need to find the squared distance between these two points using the distance formula. The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. The squared distance is simply $(x_2-x_1)^2 + (y_2-y_1)^2$.

$$(PQ)^2 = (x_Q - x_P)^2 + (y_Q - y_P)^2$$ $$(PQ)^2 = (8 - 1)^2 + (6 - (-2))^2$$ $$(PQ)^2 = (7)^2 + (6 + 2)^2$$ $$(PQ)^2 = (7)^2 + (8)^2$$ $$(PQ)^2 = 49 + 64$$ $$(PQ)^2 = 113$$

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Summary and Key Takeaways

This problem effectively tests two important concepts:

1. Concurrency of Lines: Recognizing that a linear relationship between the coefficients of a line equation ($a, b, c$) can directly determine a fixed point of concurrency. In this case, $a-2b+c=0$ implies the line $ax+by+c=0$ passes through $(1, -2)$.
2. Conditions for Infinitely Many Solutions of a System of Linear Equations: For a system of $n$ equations in $n$ variables, infinitely many solutions exist if and only if the determinant of the coefficient matrix ($D$) is zero, AND all determinants formed by replacing a coefficient column with the constant terms column ($D_x, D_y, D_z$, etc.) are also zero.

By applying these principles systematically, we found $P=(1, -2)$ and $Q=(8, 6)$, leading to a squared distance $(PQ)^2 = 113$. Always ensure careful calculation of determinants and proper substitution of found values.