The points on `x+y=4` that lie at a unit distance from the line `4x+3y-10=` are
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Let the required point to be (h,k)and point (h,k) lies on the line `x+y+4` i.e.,`h+k=4` The distance of the point (h,k) from the line `4x+3y=10` is ` |(4h+3k-10)/(sqrt(16+9))|=1` `4h+3k-10=+-5` `4h+3k=15` Taking positive sign, From Eq.(i) `h=4-k` put in Eq.(ii),we get `4(4-k)+3k=15` `rArr16-4k+3k=15` `rArr k=1` On putting `k=1` in Eq.(i),we get `h+1=rArrh=3` So,the point is `(3,1)`. Taking negative sign, `4h+3k-10=-5` `rArr 4(4-k)+3k=5` `rArr 16-4k+3k=5` `rArr -k=5-16=-11` `therefore k=11` On putting `k=11=4rArrh=-7` Hence, the required poitns are `(3,1)` and `(-7,11)`
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