JEE Advanced 2013 · Paper 1 · Mathematics Q48

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MathematicsSingle correct+2Question 48 of 60

For $a>b>c>0$ , the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and $bx+ay+c=0$ is less than $2\sqrt2$ . Then

a+b-c>0

a-b+c<0

a-b+c>0

a+b-c<0

Answer and worked solution

Official answer: A

Intersection. By symmetry it lies on $y=x$ : $(a+b)x+c=0$ , so the point is $\left(-\frac c{a+b},-\frac c{a+b}\right)$ .

Distance condition. $\sqrt2\left(1+\dfrac c{a+b}\right)<2\sqrt2$ gives $c<a+b$ , i.e. $a+b-c>0$ : (A).

Also $a>b$ and $c>0$ give $a-b+c>0$ outright, so (C) holds too. The question was printed as single-correct, and the official key accepts (A), (C) or both.

(A) (key: A or C or AC)

Topics:Distance and section formulae Equations of a line

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JEE Advanced 2013 · Paper 1, Mathematics question 48