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Question
The lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent if a, b, c are in G.P.
Options
True
False
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Solution
This statement is False.
Explanation:
Given equations are
ax + 2y + 1 = 0 .......(i)
bx + 3y + 1 = 0 ......(ii)
cx + 4y + 1 = 0 .......(iii)
Solving equation (i) and (ii) we get
ax + 2y + 1 = 0
⇒ y = `(-ax - 1)/2`
Putting the value of y in equation (ii) we have
`bx + 3((-ax - 1)/2) + 1` = 0
⇒ 2bx – 3ax – 3 + 2 = 0
⇒ (2b – 3a)x = 1
⇒ x = `1/(2b - 3a)`
∴ y = `(-a(1/(2b - 3a)) - 1)/2`
= `(-a - 2b + 3a)/(2(2b - 3a))`
= `(2a - 2b)/(2(2b - 3a))`
= `(a - b)/(2b - 3a)
So, the point of intersection of equation (i) and (ii) is
`(1/(2b - 3a), (a - b)/(2b - 3a))`
If equation (i), (ii) and (iii) are concurrent, then the above point must lie on equation (iii)
⇒ `c[1/(2b - 3a)] + 4[(a - b)/(2b - 3a)] + 1` = 0
⇒ `(c + 4a - 4b + 2b - 3a)/(2b - 3a)` = 0
⇒ c + a – 2b = 0
⇒ 2b = a + c
So, a, b and c are in A.P. and not in G.P.
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