Let the circumcentre of a triangle with verti | vedclass

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Question

$m _{ AC } \longrightarrow \infty$

$m _{ PD }=0$

$D \left(\frac{ a + a }{2}, \frac{ b +3}{2}\right)$

$D \left( a , \frac{ b +3}{2}\right)$

Vedclass · Accepted Answer

$\frac{4}{7}$

Vedclass · Answer

$m _{ AC } \longrightarrow \infty$

$m _{ PD }=0$

$D \left(\frac{ a + a }{2}, \frac{ b +3}{2}\right)$

$D \left( a , \frac{ b +3}{2}\right)$

$m _{ PD }=0$

$\frac{ b +3}{2}-1=0$

$b +3-2=0$

$b =-1$

$E \left(\frac{ b + a }{2}, \frac{5+ b }{2}\right)=\left(\frac{ af }{2}, 2\right)$

$m_{C B} \cdot m_{E P}=-1$

$\left(\frac{5-b}{b-a}\right)=\left(\frac{2-1}{\frac{a-1}{2}-1}\right)=-1$

$\left(\frac{6}{-1-a}\right)=\left(\frac{2}{a-3}\right)=-1$

$12=(1+a)(a-3)$

$12=a^{2}-3 a+a-3$

$a^{2}-2 a-15=0$

$(a-5)(a+3)=0$

$a=5$ or $a=-3$

Given $a b > 0$

$a (-1) > 0$

$- a > 0$

$a < 0$

$a=-3$ Accept

AP line $A (-3,3) P (1,1)$

$y-1=\left(\frac{3-1}{-3-1}\right)(x-1)$

$-2 y+2=x-1$

$x+2 y=3 \quad$ Appling $\ldots . .(1)$

Line $BC B(-1,5)$

$C (-3,-1)$

$( y -5)=\frac{6}{2}( x +1)$

$y-5=3 x+3$

$y=3 x+8$

Solving (1) \& (2)

$x+2(3 x+8)=3$

$7 x+16=3$

$7 x=-13$

$x=-\frac{13}{7}$

$y=3\left(-\frac{13}{7}\right)+8$

$=\frac{-39+56}{7}$

$y=\frac{17}{7}$

$x+y=\frac{-13+17}{7}=\frac{4}{7}$

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.

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