First slide

Cartesian plane

Question

Vertices of a variable triangle are $A (3, 4), B (5 \cos θ, 5 \sin θ) and C (5 \sin θ, - 5 \cos θ)$ Then, locus of its orihocenire is

Moderate

A

$(x + y - 1)^{2} + (x - y - 7)^{2} = 100$
B

$(x + y - 7)^{2} + (x - y - 1)^{2} = 100$
C

$(x + y - 7)^{2} + (x + y - 1)^{2} = 100$
D

$(x + y - 7)^{2} + (x - y + 1)^{2} = 100$

Solution

We observe that $O A = O B = O C$ . Therefore $0 (0, 0)$ is the circumcentre of $△ A B C . Let O^{'} (h, k)$ be the coordinates of its orthocentre
The centroid G $(\frac{3 + 5 \cos θ + 5 \sin θ}{3}, \frac{4 + 5 \sin θ - 5 \cos θ}{3})$
divides the line segment joining $O O^{'}$ in the ratio 1 : 2.
$∴ \frac{1 \times h + 2 \times 0}{1 + 2} = \frac{3 + 5 \cos θ + 5 \sin θ}{3}$
and
$\frac{1 \times k + 2 \times 0}{1 + 2} = \frac{4 + 5 \sin θ - 5 \cos θ}{3}$
$\overset{1}{O (0, 0)} ⟶ 2 O O' (h, k)$
$\begin{array}{l} \Rightarrow h = 3 + 5 \cos θ + 5 \sin θ and k = 4 + 5 \sin θ - 5 \cos θ \\ \Rightarrow h + k - 7 = 10 \sin θ and h - k + 1 = 10 \cos θ \\ \Rightarrow (h + k - 7)^{2} + (h - k + 1)^{2} = 100 \end{array}$
Hence, the locus of $(h, k) is (x + y - 7)^{2} + (x - y + 1)^{2} = 100$

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