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Question

# The points on the line x+y=4 lying at a unit distance from the line 4x+3yâˆ’10=0 are

A
(7,11),(3,1)
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B
(7,11),(3,1)
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C
(7,11),(3,7)
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D
(7,3),(11,7)
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E
(2,2),(3,1)
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Solution

## The correct option is B (−7,11),(3,1)Let (h,k) be the point on the line x+y=4Then, h+k=14 .... (i)Also, given thatPerpendicular distance from (h,k) to the line4x+3y−10=0 is 1.i.e., ∣∣∣4h+3k−10√16+9∣∣∣=1⇒4h+3k−10=±5⇒4h+3k=10±5⇒4h+3k=15 ...(ii)and 4h+3k=5 ... (iii)Now, on solving Eqs. (i) and (ii), we get4h+3(4−h)=15⇒h=3Then from Eq. (i), k=1Again, on solving Eqs. (i) and (iii), we get4h+3(4−h)=5⇒h=−7Then from Eq. (i), k=11So, the required points are (3,1) and (−7,11).

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