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Question

# The equation of a straight line passing through a point (−5,4) and which cuts off an intercept of √2 units between the lines x+y+1=0 and x+y−1=0 is

A
x2y13=0
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B
2xy+14=0
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C
xy+9=0
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D
xy+10=0
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Solution

## The correct option is C x−y+9=0Let the given point be A=(−5,4) and the given lines be l1→x+y+1=0 and l2→x+y−1=0The point A lies on l1 If segment AM⊥l2 and M lies on l2, then, the distance AM is given by, ⇒AM=|−5+4−1|√12+12=2√2=√2 ⇒ This means that if B is any point on l2 then AB>AM. No line other than AM cuts off an intercept of length √2 between l1 and l2. ⇒To determine the equation of AM, we need to find the co-ordinates of the Point M Since, AM⊥l2 and the slope l2 is −1, the slope ofAM must be 1. Also A(−5,4) lies on AM By the point slope formula, the equation of the required line is ⇒y−4=1(x−(−5)) ⇒y−4=x+5 ⇒x−y+9=0

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