Question
Find the value of x, if the distance between the points (x, –1) and (3, 2) is 5. [2 MARKS]
Answer: Option A
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Concept : 1 Mark
Application : 1 Mark
Let p (x , -1) and Q (3 ,2) be the given points. Then,
PQ = 5 (Given)
Distance between the points is given by
√(x1−x2)2+(y1−y2)2
⇒√(x−3)2+(−1−2)2=5
⇒(x−3)2+9=52 [Squaring both sides]
⇒x2−6x+18=25
⇒x2−6x−7=0
⇒(x−7)(x+1)=0
⇒x=7 or x=−1
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