Question

Question 13
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.

Answer 1

Since, (a,b) is the mid-point of the line segment AB.
$\therefore$ $(a,b)=(\frac{10+k}{2},\frac{-6+4}{2})$
$[\because \text{Mid-point of a line segment having points}]$
$(x_1,y_1)and(x_2,y_2)]$
$=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})]$
$\Rightarrow (a,b)=(\frac{10+k}{2},-1)$
$\text{Now, equating coordinates on both sides, we get}$
$\therefore$ $a=\frac{10+k}{2}\text{and b= -1}$ ...(i)
$\text{Given, a-2b=18}$
$\text{From Eq,(i), a -2 (-1) = 18}$
$\Rightarrow a+2=18\Rightarrow a=16$
$\text{From Eq.(i),}16=\frac{10+k}{2}$
$\Rightarrow 32=10+k\Rightarrow k=32$
$\text{Hence, the required value of k is 22.}$
$\Rightarrow k=22$
$\therefore A=(10,-6),B=(22,4)$
$\text{Now, distance between A(10,-6) and B(22,4)}$
$AB=\sqrt{(22-10{)}^2+(4+6{)}^2}$
$\left[\begin{array}{c}\\ \because \text{Distance between the points}(x_1,y_1)and(x_2,y_2),d\\ =\sqrt{(x_2-x_1{)}^2+(x_2-y_1{)}^2}\end{array}\right]$
$=\sqrt{(12{)}^2+(10{)}^2}=\sqrt{144+100}$
$=\sqrt{244}=2\sqrt{61}$
$\text{Hence, the required distance of AB is}2\sqrt{61}.$