Question

If the distance between (2,-3) and (10,x) is
10 units then find x​

Answer 1

let the points be Pand Q

PQ=√(x²-x¹)+(y²-y¹)²

10 = √(10-2)²+(x--3)²

100 = 64+x²+9+ 2×3x

x²+6x-27=0

x²-3x+9x-27=0

x(x-3)+9(x-3)=0

x+9=0 or x-3=0

x= -9 or x=3

Answer 2

Given :-

• The distance between (2,-3) and (10,x) is
• 10 units.

To Find :-

• The value of x = ?

Solution :-

The distance between the two points (x₁,y₁) and (x₂,y₂) is given by Distance formula :

• Distance = √(x₂ - x₁)² + (y₂ - y₁)²

Where,

• x₁ = 2
• x₂ = 10
• y₁ = x
• y₂ = -3
• Distance = 10 units

➻ Distance = √(x₂ - x₁)² + (y₂ - y₁)²

➻ 10 = √(10 - 2)² + (x - {-3})²

➻ 10 = √(8)² + (x + 3)²

➻ 10 = √64 + x² + 9 + 6x

➻ 10 = x² + 6x + √73

➻ 100 = x² + 6x + 73....[Squaring the both sides]

➻ x² + 6x +73 - 100

➻ x² + 6x - 27 = 0

➻ x² - 3x + 9x - 27 = 0

➻ x(x - 3) + 9(x - 3) = 0

➻ (x - 3) (x + 9) = 0

➻ x = 3, or x = -9

Therefore,the value of x = 3 or -9.