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 be a point such that the system of equations x + y + z = 6, 2x + 5y + z = and x + 2y + 3z = 4, has infinitely many solutions. The

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Solution

Q.
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 $Q (α, β)$ be a point such that the system of equations x + y + z = 6, 2x + 5y + $α$ z = $β$ and x + 2y + 3z = 4, has infinitely many solutions. Then ${(P Q)}^{2}$ is equal to __________. [2024]

Ans.
(113)

a, b, c are in A.P. $\Rightarrow$ 2b = a + c $\Rightarrow$ a – 2b + c = 0 and ax + by + c = 0 are concurrent.

Two lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ are concurrent if

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} ∴ \frac{1}{x} = \frac{- 2}{y} = \frac{1}{1} = 0$

$\Rightarrow x = 1 and y = - 2$

So, P(1, –2) is the point of concurrency.

Now, x + y + z = 6 ... (i)

2x + 5y + $α$ z = $β$ ... (ii)

x + 2y + 3z = 4 ... (iii)

has infinitely many solution.

On solving these equation, we have

x + y = 6 – z, x + 2y = 4 – 3z

On solving these two, we get –y = 2 + 2z

$\Rightarrow$ y = –2(1 + z) $\Rightarrow$ x = 6 – z + 2 + 2z $\Rightarrow$ x = 8 + z

From (ii), we get 2(8 + z) + 5(–2(1 + z)) + $α$ z = $β$

$\Rightarrow 6 - 8 z + α z = β$

$\Rightarrow z (α - 8) = β - 6$

For infinitely many solution, $α = 8, β = 6$

$| P Q |^{2}$ $= {(\sqrt{{(8 - 1)}^{2} + {(6 + 2)}^{2}})}^{2} = 113$

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