# 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 $3$.