**Question:**

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.

**Solution:**

Since, (a, b) is the mid-point of line segment AB.

$\therefore$ $(a, b)=\left(\frac{10+k}{2}, \frac{-6+4}{2}\right)$

$\left[\right.$ since, mid-point of a line segment having points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\right]$

$\Rightarrow$ $(a, b)=\left(\frac{10+k}{2},-1\right)$

Now, equating coordinates on both sides, we get

$\therefore$ $a=\frac{10+k}{2}$ ...(i)

and $b=-1$ ...(ii)

Given, $a-2 b=18$

From Eq. (ii), $\quad a-2(-1)=18$

$\Rightarrow$ $a+2=18 \Rightarrow a=16$

From Eq. (i), $16=\frac{10+k}{2}$

$\Rightarrow$ $32=10+k \Rightarrow k=22$

Hence, the required value of $k$ is 22 .

$\Rightarrow$ $k=22$

$\therefore$ $A \equiv(10,-6), B \equiv(22,4)$

Now, distance between $A(10,-6)$ and $B(22,4)$,

$A B=\sqrt{(22-10)^{2}+(4+6)^{2}}$

$\left[\because\right.$ distance between the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$

$=\sqrt{(12)^{2}+(10)^{2}}=\sqrt{144+100}$

$=\sqrt{244}=2 \sqrt{61}$

Hence, the required distance of AB is 2√61.