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

Answer:

y= -3 (or) -9

Hope you have understood

Answer 2

To Find :-

• The value of y.

Solution :-

Given,

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

Where,

• x1 = 2
• y1 = -3
• x2 = 10
• y2 = y
• PQ = 10 units

Applying distance formula,

↪ XY = √(x2 - x1)² + (y2 - y1)²

[ Put the values ]

↪ PQ = √(10 - 2)² + (y - (-3)²

↪ PQ = √(10 - 2)² + (y + 3)²

↪ 10 = √(8)² + (y + 3)²

↪ 10 = √64 + (y + 3)²

[ Squaring both side, we get ; ]

↪ 100 = 64 + (y + 3)²

↪ 100 - 64 = (y + 3)²

↪ 36 = (y + 3)²

↪ y + 3 = ±6

↪ y = 6 - 3 OR y = -6 - 3

↪ y = 3 OR y = -9

Therefore,

The value of y is 3 and -9.