Let the area of a PQR with vertices P(5, 4), Q(–2, 4) and R(a, b) be 25 square units. If its orthocenter and centroid are and C(c, d) respectively, then c + 2d is equal to [2025]

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Solution

Q.
Let the area of a $△$ PQR with vertices P(5, 4), Q(–2, 4) and R(a, b) be 35 square units. If its orthocenter and centroid are $O (2, \frac{14}{5})$ and C(c, d) respectively, then c + 2d is equal to [2025]

1 2

2 3

3 $\frac{7}{3}$

4 $\frac{8}{3}$

Ans.
(2)

Since, $P R ⊥ Q T$

$∴ \frac{b - 4}{a - 5} \times \frac{- 6 / 5}{4} = - 1 \Rightarrow \frac{b - 4}{a - 5} \times \frac{3}{10} = 1$

$\Rightarrow 10 a - 3 b = 38$ ... (i)

Also, $P S ⊥ Q R$

$∴ \frac{- 6 / 5}{- 3} \times \frac{b - 4}{a + 2} = - 1$

$\Rightarrow 5 a + 2 b = - 2$ ... (ii)

Solving equations (i) and (ii), we get

a = 2, b = - 6

Now, centroid of $△$ PQR = (c, d)

$\Rightarrow (\frac{5 - 2 + 2}{3}, \frac{4 + 4 - 6}{3}) = (c, d)$

$\Rightarrow c = \frac{5}{3}, d = \frac{2}{3}$

$∴ c + 2 d = \frac{5}{3} + \frac{4}{3} = 3$ .

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