JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Straight Lines

For

a > b > c > 0

, the distance between

(1, 1)

and the point of intersection of the lines

a x + b y + c = 0

and

b x + a y + c = 0

is less than

22

. Then

(A) $a + b - c > 0$
(B) $a - b + c < 0$
(C) $a - b + c > 0$
(D) $a + b - c < 0$

Solution

The intersection point of

a x + b y + c = 0

and

b x + a y + c = 0

is found by solving the system. Subtracting gives

(a - b) (x - y) = 0

, so

x = y

. Substituting into the first equation:

a x + b x + c = 0

, thus

x = - \frac{c}{a + b}

. The intersection point is

(- \frac{c}{a + b}, - \frac{c}{a + b})

. The distance from

(1, 1)

2 (1 + \frac{c}{a + b}) < 22

, which gives

1 + \frac{c}{a + b} < 2

, hence

\frac{c}{a + b} < 1

, so

c < a + b

, i.e.,

a + b - c > 0

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option

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