The lines $a x + 2 y + 1 = 0, b x + 3 y + 1 = 0$
and $c x + 4 y + 1 = 0$ are concurrent, show that $a, b, c$ are in $A . P .$

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Solution

As above $∣ ∣ ∣ ∣ \begin{matrix} a & 2 & 1 b & 3 & 1 c & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Apply $R_{1} + R_{3} - {2 R}_{2}$
$∣ ∣ ∣ ∣ \begin{matrix} a + c - 2 b & 0 & 0 b & 3 & 1 c & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Or $(a + c - 2 b) (3 - 4) = 0$ $∴ a + c - 2 b = 0$
Or $a + c = 2 b$ $∴ a, b, c$ are in $A . P .$

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Similar questions

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

a x + 2 y + 1 = 0

b x + 3 y + 1 = 0

c x + 4 y + 1 - 0

a, b, c

a, b, c

a x + 2 y + 1 = 0

b x + 3 y + 1 = 0

c x + 4 y + 1 = 0

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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