Question
Determine the points yz-plane equidistant from the points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).
Let Q(0, y, z) be the required point.
So
(AQ)2 = (BQ)2 ⇒ (0 - 1)2 + (y + 1)2 - (z - 0)2 = (0 - 2) + (y + 1)2 + (z - 2)2
⇒ 1 + y2 + 1 + 2y + z2 = 4 + y2 + 1 - 2y + z2 + 4 - 42
⇒ 4y + 4z = 7 ... (i)
(BQ)2 = (CQ)2 ⇒ (0 - z)2 + (y - 1)2 + (z - 2)2 = (0 - 3)2 + (y - 2)2 (2 + 1)2
⇒ 4 + y2 + 1 - 2y + z2 + 4 - 4z - 9 + y2 + 4 - 4y + z2 + 1 + 2z
⇒ 2y - 6z = 5 ... (ii)
(AQ)2 = (CQ)2 ⇒ (0 - 1)2 + (y + 1)2 + (z - 0)2 = (0 - 3)2 + (y - 2)2 (z + 1)2
⇒ 1 + y2 + 2y + 1 + z2 = 9 + y2 - 4y + 4 + z2 + 1 +2z
⇒ 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 $=\Big(0,\ \frac{31}{16},\ \frac{-3}{16}\Big)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| column I | column II | ||
| (a) | Throught the point (2, 1) is | (a) | 2x - y = 4 |
| (b) | perpendicular to the line x + 2y + 1 = 0 is | (b) | x + y - 5 = 0 |
| (c) | parpallel to the line 3x + 4y + 5 = 0 | (c) | x - y - 1 |
| (d) | Equally inlined to the axis is | (d) | 3x - 4y - 1 = 0 |
Find the sixth term in the expansion $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$ if the binomial coefficient of the term from the end is 45.
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