If Acos - sin- Bsin- -cos- C 1-2 are the vertices of a - ABC- then as - varies- the locus of its centroid is-

The correct option is C 3(x^2 + y^2) 2x 4y + 1 = 0 Ref. Image. ( L , K )≡ ( cosα #160; + sinα #160; + 13,sinα #160; cosα #160; + α3 ) On ...

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Standard XII

Mathematics

Orthocentre

If Acos α, ...

Question

If $A (c o s α, s i n α), B (sin α, - c o s α), C (1, 2)$ are the vertices of a $Δ A B C$ , then as $α$ varies, the locus of its centroid is:

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

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x^{2} + y^{2} - 2 x - 4 y + 1 = 0

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3 (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0

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None of these

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Solution

The correct option is C $3 (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0$
Ref. Image.
$(L, K) \equiv (\frac{c o s α + s i n α + 1}{3}, \frac{s i n α - c o s α + α}{3})$
On compare,
$3 h = c o s α + s i n α + 1$ ...1)
$3 k = s i n α - c o s α + α$ ...(2)
____________________
add
$3 h + 3 k = 2 s i n α + 3$
$s i n α = \frac{3 h + 3 k - 3}{2}$
on Substitute (1) & (2)
$3 h - 3 k = 2 c o s α - 1$
$c o s α = \frac{3 h - 3 k + 1}{2}$
$s i n^{2} α + c o s^{2} α = 1$
${(\frac{3 h + 3 k - 3}{2})}^{2} + {(\frac{3 h - 3 k + 1}{2})}^{2} = 1$
$\frac{9 h^{2} + 9 k^{2} + 9 + 2 (9 h k - 9 k - 9 h)}{4} + \frac{9 h^{2} + 9 k^{2} + 1 + 2 - (3 h k - 3 k + 3 h)}{4}$
$18 h^{2} + 18 k^{2} + 18 h k - 18 k - 18 h - 18 h k - 6 k + 6 h + 10 = 4$
$18 h^{2} + 18 k^{2} - 24 k - 12 k + 6 = 0$
$3 h^{2} + 3 k^{2} - 4 k - 2 h + 1 = 0$
$3 (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0$

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If A(cosα,sinα),B(sinα,−cosα),C(1,2) are the vertices of a ΔABC, then as α varies, the locus of its centroid is:

If $A (c o s α, s i n α), B (sin α, - c o s α), C (1, 2)$ are the vertices of a $Δ A B C$ , then as $α$ varies, the locus of its centroid is: