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

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Given:

$(a, b)$ is the mid-point of the line segment joining the points $A (10, -6), B (k, 4)$ and $a – 2b = 18$.

To do:

We have to find the value of $k$ and the distance AB.

Solution:

We know that,

Mid-point of a line segment joining points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1+y_1}{2}, \frac{y_1+y_2}{2})$

Therefore,

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

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

Equating coordinates on both sides, we get,

$a=\frac{10+k}{2}$ and $b=-1$

Given,

$a-2 b=18$

This implies,

$a-2(-1)=18$

$a+2=18$

$a=18-2=16$

$16=\frac{10+k}{2}$

$16(2)=10+k$

$k=32-10=22$

Distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2}) =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Distance between the points $\mathrm{A}(10,-6)$ and $\mathrm{B}(22,4)$

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

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

$=\sqrt{144+100}$

$= \sqrt{244}$

$=2 \sqrt{61}$

Therefore, the value of $k$ is $22$ and the distance $\mathrm{AB}$ is $2 \sqrt{61}$.

Updated on 10-Oct-2022 13:28:51