the length of a line segment is of 10 units n 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.

Hi..!!!

Here is the answer to your question.

Let AB be the line segment of length 10 units, where A ≡ (2, –3). Let ordinate of B be 'm', then B ≡ (10, m).

Now, AB = 10

⇒ AB² = 100

⇒ (2 – 10)² + (–3  –m)² = 100

⇒ (8)² + (3 + m)² = 100

⇒ 64 + 9 + m² + 6m = 100

⇒ m² + 6m – 27 = 0

⇒ m² – 3m + 9m – 27 = 0

⇒ m (m – 3) + 9 (m –3) = 0

⇒ (m  + 9)  (m –3) = 0

⇒ m  = –9, 3

Thus, point B can be (10, –9) or (10, 3).

Cheers..!!!