Find all possible values of × for which the distance between the points A(x, – 1) and B(5, 3) is 5 units.
Given:
Distance AB = 5 units
By distance formula, as shown below:
AB = √{(5 – x)2 + (3 – (– 1))2}
5 = √{(5 – x)2 + (4)2}
5 = √{25 + x2 – 10x + 16}
5 = √{41 + x2 – 10x}
Squaring both sides we get
25 = 41 + x2 – 10x
⇒ 16 + x2 – 10x = 0
⇒ (x – 8)(x – 2) = 0
⇒ × = 8 or × = 2
∴ The values of × can be 8 or 2
3