JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 9)
Let the circumcentre 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$$\left(k_{1}, k_{2}\right)$$, then $$k_{1}+k_{2}$$ is equal to :
2
$$\frac{4}{7}$$
$$\frac{2}{7}$$
4
Explanation
_29th_July_Morning_Shift_en_9_1.png)
Let D be mid-point of AC, then
$${{b + 3} \over 2} = 1 \Rightarrow b = - 1$$
Let E be mid-point of BC,
$${{5 - b} \over {b - a}}\,.\,{{{{(3 + b)} \over 2}} \over {{{a + b} \over 2} - 1}} = - 1$$
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 $$BC \Rightarrow y = 3x + 8$$
Line $$AP \Rightarrow y = {{3 - x} \over 2}$$
Point of intersection $$\left( {{{ - 13} \over 7},{{17} \over 7}} \right)$$
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