If the three lines ax + a 2 y + 1 = 0, bx + b 2 y + 1 = 0 and cx + c 2 y + 1 = 0 are concurrent, show that at least two of three constant a, b, c are equal.

Question

Philoid · Accepted Answer

Given:

ax + a²y + 1 = 0

bx + b²y + 1 = 0

cx + c²y + 1 = 0

To prove:

At least two of three constant a, b, c are equal.

Concept Used:

If three lines are concurrent then determinant of equation is zero.

Explanation:

The given lines can be written as follows:

ax + a²y + 1 = 0 … (1)

bx + b²y + 1 = 0 … (2)

cx + c²y + 1 = 0 … (3)

The given lines are concurrent.

∴

Applying the transformation R₁→R₁-R₂ and R₂→R₂-R₃:

⇒

⇒ (a – b)(b – c)(c – a) = 0

⇒ a – b = 0 or b – c = 0 or c – a = 0

⇒ a = b or b = c or c = a

Hence proved, atleast two of the constants a,b,c are equal .

If the three lines ax + a2y + 1 = 0, bx + b2y + 1 = 0 and cx + c2y + 1 = 0 are concurrent, show that at least two of three constant a, b, c are equal.

If the three lines ax + a²y + 1 = 0, bx + b²y + 1 = 0 and cx + c²y + 1 = 0 are concurrent, show that at least two of three constant a, b, c are equal.