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.
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.
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