Find the values of y for which the distance between the points P (2, -

1.	Find the values of y for which the distance between the points P (2, - 3) and Q (10, y) is 10 units.
Answer» Given: the distance between the points P (2, -3) and Q (10, y) is 10 units. To find: The value of y. Solution: Coordinates are P (2, - 3) and Q (10, y) We use distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to find the distance between two points. Since PQ = 10 units S0, \(\sqrt{(10 - 2)^2 + (y + 3)^2}\) = 10 On squaring both sides, we get (10 - 2)² + (y + 3)²= 100 ⇒ 82 + (y + 3)² = 100 ⇒ 64 + y² + 6y + 9 = 100 ⇒ 73 + y² + 6y = 100 ⇒ 73 + y² + 6y -100 = 0 ⇒ y² + 6y - 27 = 0 In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized. ⇒ y² + 9y - 3y - 27 = 0 ⇒y(y + 9) - 3(y + 9) = 0 ⇒(y - 3)(y + 9) = 0 ⇒y = 3, - 9

Find the values of y for which the distance between the points P (2, - 3) and Q (10, y) is 10 units.

Answer»

Given:

the distance between the points P (2, -3) and Q (10, y) is 10 units.

To find:

The value of y.

Solution:

Coordinates are P (2, - 3) and Q (10, y)

We use distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to find the distance between two points.

Since PQ = 10 units

S0,

\(\sqrt{(10 - 2)^2 + (y + 3)^2}\) = 10

On squaring both sides, we get

(10 - 2)² + (y + 3)²= 100

⇒ 82 + (y + 3)² = 100

⇒ 64 + y² + 6y + 9 = 100

⇒ 73 + y² + 6y = 100

⇒ 73 + y² + 6y -100 = 0

⇒ y² + 6y - 27 = 0

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ y² + 9y - 3y - 27 = 0

⇒y(y + 9) - 3(y + 9) = 0

⇒(y - 3)(y + 9) = 0

⇒y = 3, - 9

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