Question

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

Answer 1

Step-by-step explanation:

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Answer 2

Answer :-

• Value of y is 9 or -3.

Given :-

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

To Find :-

• Value of y.

Solution :-

Here

• x1 = 2
• x2 = 10
• y1 = -3
• y2 = y

As we know that

For finding distance between two points, the formula is

• √[(x2 - x1)² + (y2 - y1)²

Now

⇒ PQ = √[(10 - 2)² + (y + 3)²]

⇒ 10 = √[(8)² + y² + 6y + 9]

⇒ 10 = √[64 + y² + 6y + 9]

⇒ 10 = √[64 + 9 + 6y + y²]

⇒ (10)² = 73 + 6y + y²

⇒ 100 - 73 = 6y + y²

⇒ 27 = 6y + y²

⇒ y² + 6y - 27 = 0 [ it's in the form of quadratic equation ]

⇒ y² + 3y - 9y - 27 = 0 [ split the middle term ]

⇒ y (y + 3) - 9 (y + 3)

⇒ (y - 9) (y + 3)

⇒ y = 9,-3

Hence, the value of y is 9 or -3.