If a, b, c are in A.P., prove that the straight lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent.

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Solution

The given lines can be written as follows:

ax + 2y + 1 = 0 ... (1)

bx + 3y + 1 = 0 ... (2)

cx + 4y + 1 = 0 ... (3)

Consider the following determinant.
$|\begin{matrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{matrix}|$

Applying the transformation $R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3}$ ,

$|\begin{matrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{matrix}| = |\begin{matrix} a - b & - 1 & 0 \\ b - c & - 1 & 0 \\ c & 4 & 1 \end{matrix}|$

$\Rightarrow |\begin{matrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{matrix}| = (- a + b + b - c) = 2 b - a - c$

Given:
2b = a + c

$|\begin{matrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{matrix}| = a + c - a - c = 0$

Hence, the given lines are concurrent, provided 2b = a + c.

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Similar questions

a x + 2 y + 1 = 0, b x + 3 y + 1 = 0

c x + 4 y + 1 = 0

a, b, c

A . P .

a, b, c

a x + 2 y + 1 = 0

b x + 3 y + 1 = 0

c x + 4 y + 1 = 0

a x + 2 y + 1 = 0

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c x + 4 y + 1 - 0

a, b, c

if the lines ax+2y+1=0, bx+3y+1=0, cx+4y+1=0 are concurrent then a,b,c are in:

a) AP

b)GP

c)HP

d)Neither GP or AP

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