The length of a line segment is of 10 units and the coordinates of one end-point are $(2, -3)$. If the abscissa of the other end is 10, find the ordinate of the other end.



Given:

The length of a line segment is of 10 units and the coordinates of one end-point are $(2, -3)$ and the abscissa of the other end is 10.

To do:

We have to find the ordinate of the other end.

Solution:

 Let the ordinate of the other end be $y$.

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

The distance between $(2, -3)$ and $(10, y)$ is,

\( \Rightarrow \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=10 \)

\( \Rightarrow \sqrt{(10-2)^{2}+(y+3)^{2}}=10 \)

\( \Rightarrow(8)^{2}+(y+3)^{2}=100 \)

\( \Rightarrow 64+y^{2}+6 y+9=100 \)

\( \Rightarrow y^{2}+6 y+73-100=0 \)

\( \Rightarrow y^{2}+6 y-27=0 \)

\( \Rightarrow y^{2}+9 y-3 y-27=0 \)

\( \Rightarrow  y(y+9)-3(y+9)=0 \)

\( \Rightarrow  (y+9)(y-3)=0 \)

\( y+9=0 \) or \( y-3=0 \)

\( y=-9 \) or \( y=3 \)

Therefore, the ordinate of the other end is $-9$ or $3$.

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