If (a,b) is the mid - point of the line segment joining the points A (10,-6), B(k,4) and a-2b =18, then find the value of k and the distance AB.

Explanation

Since , (a,b) is the mid- point of line segment AB.
`:.(a,b)=((10+k)/(2),(-6+4)/(2))`
[ since ,mid point of a line segment having points `(x_(1),y_(1)) and (x_(2),y_(2))=((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))]`
`rArr(a,b)=((10+k)/(2),-1)`
Now , equating coordinates on both sides , we get
`:. a=(10+k)/(2)` " " ...(i)
and `b=-1` " " ...(ii)
Given , a-2b=18
From Eq. (ii) , `a-2(-1)=18`
`rArr a+2=18rArra=16`
From Eq. (i),`16=(10+k)/(2)`
`rArr32=10+krArrk=22`
Hence , the required value of k is 22.
`rArrk=22`
`:.` A `-=(10,-6),B-=(22,4)` Now , distance between A (10,-6) and B (22,4),
AB`=sqrt((22-10)^(2)+(4+6)^(2))`
[`:'` distance between the points `(x_(1),y_(1))and (x_(2),y_(2))`, d `=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]`
`=sqrt((12)^(2)+(10)^(2))=sqrt(144+100)`
`=sqrt(244)=2 sqrt(61)`
Hence , the required distance of AB is `2sqrt(61)`.