Let the circumcenter of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab 0 be P(1, 1). If the line AP intersects the line BC at the point Q(k?, k?), then k? + k? is equal to

Question

Let the circumcenter of a triangle with vertices $A (a, 3), B (b, 5)$ and $C (a, b), a b > 0$ be P(1, 1). If the line AP intersects the line BC at the point $Q (k_{1}, k_{2})$ , then $k_{1} + k_{2}$ is equal to

Smart Achievers · Accepted Answer

(2)

$slope of A C = \infty$

$Slope of P D = 0$

$D (\frac{a + a}{2}, \frac{b + 3}{2}) = D (a, \frac{b + 3}{2})$

$\frac{b + 3}{2} - 1 = 0$

$b + 3 - 2 = 0$

$\Rightarrow b = - 1$

$b = - 1$

$E (\frac{b + a}{2}, \frac{5 + b}{2}) = (\frac{a - 1}{2}, 2)$

$slope of B C \times slope of E P = - 1$

$(\frac{5 - b}{b - a}) \times (\frac{2 - 1}{\frac{a - 1}{2} - 1}) = - 1$

$\Rightarrow (\frac{6}{- 1 - a}) \times (\frac{2}{a - 3}) = - 1$

$\Rightarrow 12 = (1 + a) (a - 3)$

$\Rightarrow 12 = a^{2} - 3 a + a - 3$ $\Rightarrow a^{2} - 2 a - 15 = 0$

$a = - 3 accept$

$Equation of A P, A (- 3, 3), P (1, 1)$

$y - 1 = (\frac{3 - 1}{- 3 - 1}) (x - 1)$ $\Rightarrow x + 2 y = 3$

$Equation of B C, B (- 1, 5), C (- 3, - 1)$

$y + 1 = (\frac{5 + 1}{- 1 + 3}) (x + 3)$ $= 3 x + 9$

$3 x - y + 8 = 0$

$Q \equiv (\frac{- 13}{7}, \frac{17}{7})$

Solution

Q.
Let the circumcenter of a triangle with vertices $A (a, 3), B (b, 5)$ and $C (a, b), a b > 0$ be P(1, 1). If the line AP intersects the line BC at the point $Q (k_{1}, k_{2})$ , then $k_{1} + k_{2}$ is equal to

1 $2$

2 $\frac{4}{7}$

3 $\frac{2}{7}$

4 $4$

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