#### Question

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.

#### Solution

The distance *d* between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula

d = sqrt((x_1, x_2)^2 + (y_1 - y_2)^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 + 6y`

`27 = y^2 + 6y`

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

`y^2 + 6y - 27 = 0`

`y^2 + 9y - 3y - 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 -9 or 3