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Question

The distance of the point (1, 1) from the line 2x3y4=0 in the direction of the line x+y=1, is

A
2
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B
52
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C
12
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D
12
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Solution

The correct option is A 2
Equation of line parallel to x+y=1 is x+y+c=0
Let it passes through (1,1) so the line is along direction of x+y=1
1+1+c=0c=2x+y2=0.....(i)
Given line is 2x3y4=0....(ii)
Solving (i) and (ii)
x=2,y=0
So the point of intersection is (2,0)
So the distance of (1,1) from 2x3y4=0 along x+y+1=0 is
=(21)2+(01)2=2
So option A is correct



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