Let (h,k) be the point on the line x+y=4 Then, h+k=4.....(i) Also, given that Perpendicular distance from (h,k) to the line 4x+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 get 4h+3(4−h)=15⇒h=3 Then from Eq. (i), k=1 Again, on solving Eqs. (i) and (iii), we get 4h+3(4−h)=5⇒h=−7 Then from Eq. (i), k=11 So, the required points are (3,1) and (−7,11) .