Find the point in yz−plane which is equidistant from the points A(3,2,−1),B(1,−1,0) and C(2,1,2).
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Given points are A(3,2,−1), B(1,−1,0) and C(2,1,2).
We know that x coordinate of every point in yz− plane is zero.
Any point in the yz− plane is of the form P(0,y,z).
Now, according to question
∣AP∣ = ∣BP∣ = ∣CP∣ .......... Given
⇒ AP2 = BP2
⇒ (0 − 3)2 + (y − 2)2 + (z + 1)2 = (0 − 1)2 + (y + 1)2 + (z − 0)2
⇒ 3y − z − 6 = 0 ....... (i)
and BP2 = CP2.
⇒ (0 − 1)2 + (y + 1)2 + (z − 0)2 = (0 − 2)2 + (y − 1)2 + (z − 2)2
⇒ 4y + 4z − 7 = 0 ......... (ii)
On solving, (i) and (ii) we get