jee-advanced 2013 Q48

jee-advanced · India · paper1 Straight Lines & Coordinate Geometry Point-to-Line Distance Computation

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 $2 \sqrt { 2 }$. Then
(A) $a + b - c > 0$
(B) $a - b + c < 0$
(C) $a - b + c > 0$
(D) $a + b - c < 0$

A or C or AC

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 $2 \sqrt { 2 }$. Then\\
(A) $a + b - c > 0$\\
(B) $a - b + c < 0$\\
(C) $a - b + c > 0$\\
(D) $a + b - c < 0$

Paper Questions

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