Find the point in y z - p l a n e which is equidistant from the points A \left(\right. 3 , 2 , - 1 \left.\right) , B \left(\right. 1 , - 1 , 0 \left.\right) and C \left(\right. 2 , 1 , 2 \left.\right).

Engineering

Mathematics

Line of Greatest Slope

Question

Find the point in

y z - p l a n e

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

y z -

plane is zero.

Any point in the

y z -

plane is of the form

P (0, y, z)

Now, according to question

∣ A P ∣ = ∣ B P ∣ = ∣ C P ∣

.......... Given

\Rightarrow A P^{2} = B P^{2}

\Rightarrow (0 - 3)^{2} + (y - 2)^{2} + (z + 1)^{2} = (0 - 1)^{2} + (y + 1)^{2} + (z - 0)^{2}

\Rightarrow 3 y - z - 6 = 0

....... (i)

and

B P^{2} = C P^{2}

\Rightarrow (0 - 1)^{2} + (y + 1)^{2} + (z - 0)^{2} = (0 - 2)^{2} + (y - 1)^{2} + (z - 2)^{2}

\Rightarrow 4 y + 4 z - 7 = 0

......... (ii)

On solving, (i) and (ii) we get

y = \frac{31}{16}

and

z = \frac{- 3}{16}

Hence, the required point is

(0, \frac{31}{16}, \frac{- 3}{16})

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