Determine the points yz-plane equidistant from the points A(1, -1…

Question

Determine the points yz-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$

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Answer

Let $Q(0, y, z)$ be the required point.
So
$(AQ)^2 = (BQ)^2$
$\Rightarrow (0 - 1)^2 + (y + 1)^2 - (z - 0)^2 = (0 - 2) + (y + 1)^2 + (z - 2)^2$
$\Rightarrow 1 + y^2 + 1 + 2y + z^2 = 4 + y^2 + 1 - 2y + z^2 + 4 - 42$
$\Rightarrow 4y + 4z = 7 ... (i)$
$(BQ)^2 = (CQ)^2$
$\Rightarrow (0 - z)^2+ (y - 1)^2 + (z - 2)^2 = (0 - 3)^2 + (y - 2)^2 (2 + 1)^2$
$\Rightarrow 4 + y^2+ 1 - 2y + z^2+ 4 - 4z - 9 + y^2+ 4 - 4y + z^2+ 1 + 2z$
$\Rightarrow 2y - 6z = 5 ... (ii)$
$(AQ)^2= (CQ)^2$
$\Rightarrow (0 - 1)^2 + (y + 1)^2+ (z - 0)^2= (0 - 3)^2 + (y - 2)^2 (z +^1)^2$
$\Rightarrow 1 + y^2 + 2y + 1 + z^2 = 9 + y^2 - 4y + 4 + z^2+ 1 +2z$
$\Rightarrow 6y - 2z = 12 ... (iii)$
Solving equation (i) and (ii) we get
$\text{z}=\frac{-3}{16}$ and $\text{y} = \frac{31}{16}$
Put the value of y and z in equation (iii)
$6y - 2z = 12 = 12$
$6\Big(\frac{31}{16}\Big)-2\Big(\frac{-3}{16}\Big) = 12$
$\frac{186}{16}+\frac{6}{16}=12$
$\frac{192}{16}=12$
$12=12$
$LHS = RHS.$
so,
$\text{y}=\frac{31}{16},\ \text{z}=\frac{13}{16}$
Required point $=\Big(0,\ \frac{31}{16},\ \frac{-3}{16}\Big)$

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