Question

The length of a line segment AB is 10 units. If the coordinates of one extremity are (2,-3) and the abscissa of the other extremity is 10 then the sum of all possible values of the ordinate of the other extremity is -

• A. 3
• B. -4
• C. 12
• D. -6

Answer 1

Answer: (D)

-6

Answer 2

The problem asks us to find the sum of all possible values of the ordinate (y-coordinate) of an extremity of a line segment, given the length of the segment, the coordinates of the other extremity, and the abscissa (x-coordinate) of the extremity in question.

Let the coordinates of the first extremity be $A(x_1, y_1) = (2, -3)$.
Let the coordinates of the second extremity be $B(x_2, y_2)$.
We are given that the abscissa of the second extremity is 10, so $x_2 = 10$. Let its ordinate be $y$. So, $B(10, y)$.
The length of the line segment AB is given as 10 units.

We use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is:
$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substitute the given values into the distance formula:
$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$
$10 = \sqrt{(8)^2 + (y + 3)^2}$

To eliminate the square root, square both sides of the equation:
$10^2 = (8)^2 + (y + 3)^2$
$100 = 64 + (y + 3)^2$

Now, isolate the term $(y + 3)^2$:
$(y + 3)^2 = 100 - 64$
$(y + 3)^2 = 36$

Take the square root of both sides. Remember to consider both positive and negative roots:
$y + 3 = \pm \sqrt{36}$
$y + 3 = \pm 6$

This gives us two possible cases for the value of $y$:

Case 1: $y + 3 = 6$
$y = 6 - 3$
$y = 3$

Case 2: $y + 3 = -6$
$y = -6 - 3$
$y = -9$

So, the possible values for the ordinate of the other extremity are 3 and -9.

The question asks for the sum of all possible values of the ordinate.
Sum = $3 + (-9)$
Sum = $3 - 9$
Sum = $-6$