Question

Consider a line L: ax + by + c = 0; where a, b, c are the
second, fifth and seventh terms of a non-constant
A.P. Then line L always passes through the point​

Answer 1

The line L passes through the point (2,-3)

Given, a,b,c are in AP

• First term = a = T1, Third term = b = T2, Seventh term = c = T7

Hence,

• Tn = a + (n-1)d

So,

• T1 = a + (1-1)d = a                    {verified}
• T3 = a + (3-1)d = a+2d   ⇒   b = a + 2d
• T7 = a + (7-1)d = a+6d   ⇒   c = a + 6d

Hence, equation of line changes to → ax + (a+2d)y + (a+6d) = 0        (1)

To prove that the line passes through a point, we prove L1 + λL2 = 0

Upon solving the equation (1),

•    ax + (a+2d)y + (a+6d) = 0
• ⇒ a (x+y+1) + d (2y+6) = 0
• ⇒ (x+y+1) +(d/a)(2y+6) = 0

comparing this equation to L1 + λL2 = 0,

• λ = d/a,
• L1 = x+y+1,                             (2)
• L2 = 2y + 6                           (3)

Equating (2) and (3) to zero,

•    x + y + 1 = 0    ;     2y + 6 = 0
• ⇒ x + y + 1 = 0    ;     y = -3
• ⇒ x - 3 + 1 = 0     ;      y = -3
• ⇒ x = 2, y = -3

Hence, the point is (2,-3)