Q3 of 157 Page 724

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


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