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.​

Answer 1

ANSWER

LET THE ORDINATE IF THE OTHER SIDE IS

"Y"

NOW THE COORDINATE OF THE OTHER END IS

(10,Y)

NOW THE DIATANCE IS

$\sqrt{(2 - 10) {}^{2} + ( - 3 - y) {}^{2} } = 10 \\ = > 64 + (3 + y) {}^{2} = 100 \\ = >( 3 + y) {}^{2} = 36 \\ = > 3 + y = 6 \\ = > y = 3$

therefore the ordinate is 3

Answer 2

Given:-

• 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,

To Find:-

• Find the ordinate of the other end.

Solutions:-

The distance d between two points (x1, y1) and (x2, y2) is given by the formula.

• d => √(x1 - x2)² + (x2 - y2)²

The one end of a line segment has co - ordinary (2, -3).

The abscis6 of the other end of the line segment is to be 10.

Let the ordinary of point be y.

So, the co - ordinary of the other end of the line segment is (10, y).

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

=> d = √(2 - 10)² + (-3 - y)²

=> 10 = √(-8)² + (-3 - y)²

Squaring on both sides of the equation . We have,

=> 100 = (-8)² + (-3 -y)²

=> 100 = 64 + 9 + y² + 6y

=> 27 = y² + 6y

We have a quadratic equation for y.

=> y² + 6y - 27 = 0

=> y² + 9y - 3y - 27 = 0

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

=>(y - 3) (y + 9)

The roots of the above equation are -9 and 3.

Hence, the ordinary of the other end of the line segment be -9 or 3.