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Introduction To 3D Coordinate Geometry — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsIntroduction To 3D Coordinate Geometry5 Marks
Question
Determine the points zx-plane equidistant from the points $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
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Answer

Let $R(x, 0, z)$ be the required point.
So
$(AR)^2 = (BR)^2 \Rightarrow (1 - x)^2 + (-1 - 0)^2 + (0 - z)^2 = (2 - x) + (1 - 0)^2 + (2 - z)^2$
$\Rightarrow 1 + x^2 - 2x + 1 + z^2 = 4 + x^2 - 4x + 1 + z^2 + 4z$
$\Rightarrow 2x + 4z = 7 ... (i)$
$(BR)^2 = (CR)^2 \Rightarrow (z - z)^2+ (1 - 0)^2 + (2 - z)^2 = (3 - x)^2 + (2 - 0)^2 (-1 - z)^2$
$\Rightarrow 4 + x^2- 4x + 4 + z^2- 4z = 9 + x^2- 6x + 4 + 1^+ z^2 + 2z$
$\Rightarrow 2x - 6z = 5 ... (ii)$
$(AR)^2= (CR)^2 \Rightarrow (1 - x)^2 + (1 - 0)^2+ (0 - z)^2= (3 - x)^2 + (2 - 0)^2 + (-1 - z)^2$
$\Rightarrow 1 + x^2 - 2x + 1 + z^2 = 9 + 6x + 4 + 1 + z^2$
$\Rightarrow 4x - 2z = 12 ... (iii)$
Solving equation (i) and (ii) we get
$\text{z}=\frac{1}{5},\ \text{x}=\frac{31}{10}$
Put the value of x and z in equation (iii)
4x - 2z = 12
$4\Big(\frac{31}{10}\Big)-2\Big(\frac{1}{5}\Big) = 12$
$\frac{124}{10}+\frac{2}{10}=12$
$\frac{120}{10}=12$
$12=12$
LHS = RHS.
so,
$\text{x}=\frac{31}{10},\ \text{z}=\frac{1}{5}$
Required point $=\Big(\frac{31}{10},\ 0,\ \frac{1}{5}\Big)$

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