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

Answer

Let the point on xy-plane be $P(x, y, 0).$
Now P is equidistance from $A(1, -1, 0), B(2, 1, 2)$ and $C(3, 2, -1).$
So, $AP - BP - CP$
Now,
$(AP)^2= (x - 1)^2 + (y + 1)^2 + (0 - 0)^2$
$(BP)^2 = (x - 2)^2 + (y - 1)^2 + (0 - 2)^2$
$(CP)^2 = (x - 3)^2 + (y - 2)^2 + (0 + 1)^2$
$(AP)^2 = (BP)^2$
$\Rightarrow (x - 1)^2 + (y + 1)^2 - (x - 2)^2 + (y - 1)^2 + 4$
$\Rightarrow x^2 + 1 - 2 + y^2 + 1 + 2y + z^2 - x^2 + 4 - 4x + y^2 + 1 - 2y + 4$
$\Rightarrow 2x + 4y = 7 ... (i)$
$(BP)^2 = (CP)^2$
$\Rightarrow (x - 2)^2+ (y - 1)^2 + 4 = (x - 3)^2 + (y - 2)^2 + 1$
$\Rightarrow x^2+ 4 - 4x + y^2 + 1 - 2y + z^2 + 4 = x^2 + 9 - 6x + y^2 + 4 - 4y +1$
$\Rightarrow 2x + 2y = 5 ... (ii)$
$(AP)^2= (CP)^2$
$\Rightarrow (x - 1)^2 + (y + 1)^2+ (x - 3)^2+ (y - 2)^2+^1$
$\Rightarrow x^2+ 1 - 2x + y^2 + 1 + 2y = x^2 + 9 - 6x + y^2 + 4 - 4y + 1$
$\Rightarrow 4x + 5y = 12 ... (iii)$
Solving equation (i) and (ii) we get
y = 1, $\text{x} = \frac{3}{2}$
Put x and y in equation (iii)
$4\Big(\frac{3}{2}\Big) + 6(1) = 12$
$12-12$
So, the required point is $\Big(\frac{3}{2},\ 1,\ 0\Big)$

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