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

Academic Mathematics NCERT Class 10

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 \( 2 \sqrt{61} \).

Simply Easy Learning

Updated on: 10-Oct-2022

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