Find the point in yz-plane which is equidistant from the points

Question:

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).

 

Solution:

The general point on yz plane is D(0, y, z). Consider this point is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).

$\therefore \mathrm{AD}=\mathrm{BD}$

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

Squaring both sides,

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

$9+y^{2}-4 y+4+z^{2}+2 z+1=1+y^{2}+2 y+1+z^{2}$

$-6 y+2 z+12=0$ ….(1)

Also, AD = CD

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

Squaring both sides,

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

$9+y^{2}-4 y+4+z^{2}+2 z+1=4+y^{2}-2 y+1+z^{2}-4 z+4$

$-2 y+6 z+5=0$ ..........(2)

Simultaneously solving equation (1) and (2) we get

$Y=31 / 16, z=-3 / 16$

The point which is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2) is (0, 31/16, -3/16).

 

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