The length of a line segment is of 10 units and the coordinates

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

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

Here it is given that one end of a line segment has co−ordinates (2,−3). The abscissa of the other end of the line segment is given to be 10. Let the ordinate of this point be ‘y’.

So, the co−ordinates of the other end of the line segment is (10, y).

The distance between these two points is given to be 10 units.

Substituting these values in the formula for distance between two points we have,

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

$10=\sqrt{(-8)^{2}+(-3-y)^{2}}$

Squaring on both sides of the equation we have,

$100=(-8)^{2}+(-3-y)^{2}$

$100=64+9+y^{2}+6 y$

$27=y^{2}+6 y$

We have a quadratic equation for ‘y’. Solving for the roots of this equation we have,

$y^{2}+6 y-27=0$

$y^{2}+9 y-3 y-27=0$

$y(y+9)-3(y+9)=0$

$(y+9)(y-3)=0$

The roots of the above equation are ‘−9’ and ‘3’

Thus the ordinates of the other end of the line segment could be.