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A point P (2,_1) is equal distance from the point (A,7) and (_3,A) , find A |
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Answer» ong>Step-by-step explanation:P> We know that the distance between the two points (x 1
,y 1
) and (x 2
,y 2
) is d= (x 2
−x 1
) 2 +(y 2
−y 1
) 2
Let the given points be A=(a,7) and B=(−3,a) and the third point given is P(2,−1). We first FIND the distance between P(2,−1) and A=(a,7) as FOLLOWS: PA= (x 2
−x 1
) 2 +(y 2
−y 1
) 2
= (a−2) 2 +(7−(−1)) 2
= (a−2) 2 +(7+1) 2
= (a−2) 2 +8 2
= (a−2) 2 +64
Similarly, the distance between P(2,−1) and B=(−3,a) is: PB= (x 2
−x 1
) 2 +(y 2
−y 1
) 2
= (−3−2) 2 +(a−(−1)) 2
= (−5) 2 +(a+1) 2
= 25+(a+1) 2
Since the point P(2,−1) is equidistant from the points A(a,7) and B=(−3,a), THEREFORE, PA=PB that is: (a−2) 2 +64
= 25+(a+1) 2
⇒( (a−2) 2 +64
) 2 =( 25+(a+1) 2
) 2
⇒(a−2) 2 +64=25+(a+1) 2
⇒(a−2) 2 −(a+1) 2 =25−64 ⇒(a 2 +4−4a)−(a 2 +1+2a)=−39(∵(a−b) 2 =a 2 +b 2 −2ab,(a+b) 2 =a 2 +b 2 +2ab) ⇒a 2 +4−4a−a 2 −1−2a=−39 ⇒−6a+3=−39 ⇒−6a=−39−3 ⇒−6a=−42 ⇒a= −6 −42
=7 Hence, a=7. |
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