Question:
A line segment is of length 10 units. If the coordinates of its one end are (2, −3) and the abscissa of the other end is 10, then its ordinate is
(a) 9, 6
(b) 3, −9
(c) −3, 9
(d) 9, −6
Solution:
It is given that distance between $\mathrm{P}(2,-3)$ and $\mathrm{Q}(10, y)$ is 10 .
In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,
$\mathrm{AB}^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$
So,
$10^{2}=(10-2)^{2}+(y+3)^{2}$
On further simplification,
$(y+3)^{2}=36$
$y=-3 \pm 6$
$=-9,3$
We will neglect the negative value. So,
$y=-9,3$
So the answer is (b)