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 AP intersects the line $BC$ 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

Answer: (B)

$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) $
$m _{ CB } \cdot m _{ EP }=-1$
$ \left(\frac{5-b}{b-a}\right)=\left(\frac{2-1}{a-1}\right)=-1 $
$\left(\frac{6}{-1-a}\right)=\left(\frac{2}{a-3}\right)=-1$
$12=(1+a)(a-3) $
$12=a^2-3 a+a-3$
$\Rightarrow a^2-2 a-15=0 $
$ (a-5)(a+3)=0 $
$a=5 \text { or } a=-3 $
$ \text { Given } a b>0 $
$ a(-1) >0 $
$ -a >0$
$ a< 0$
$ a =-3 \text { Accept }$
$\text { AP line A }(-3,3) P (1,1) $
$ y-1=\left(\frac{3-1}{-3-1}\right)(x-1) $
$ -2 y +2= x -1 $
$ \Rightarrow x+2 y=3 \text { Appling }......$(1)
$ \text { Line } B C B(-1,5) $
$ \text { C }(-3,-1) $
$ (y-5)=\frac{6}{2}(x+1) $
$ y-5=3 x+3 $
$ y=3 x+8 $....(2)
Solving (1) & (2)
$x+2(3 x+8)=3$
$ \Rightarrow 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} $