Question

Let the circumcentre 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 \( A P \) intersects the line \( B C \) at the point \( Q\left(k_{1}, k_{2}\right) \), then \( k_{1}+ \) \( k_{2} \) is equal to :

(a) 2

(b) \( \frac{4}{7} \)

(c) \( \frac{2}{7} \)

(d) 4

Answer 1

The correct option is (b) \(\frac{4}{7}\)

Let D be mid-point of AC, then

\(\frac{b + 3}{2} = 1\)

b = -1

Let E be mid-point of BC,

\(\frac{5-b}{b-a} . \frac{\frac{3+b}{2}}{\frac{a+b}{2} -1} = -1\)

On putting b = -1, we get a = 5 or -3

But a = 5 is rejected as ab > 0

A(-3, 3), B(-1, 5), C(-3, -1), P(1,1)

Line BC ⟹ y = 3x + 8

Line AP ⟹ y = \(\frac{3-x}{2}\)

Point of intersection \((\frac{-13}{7} , \frac{17}{7})\)