Question

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 :

• A. 2
• B. $\frac{4}{7}$
• C. $\frac{2}{7}$
• D. 4

Answer 1

Answer: (B)

$m_{PD}=0$
$D\left(\frac{a+a}{2},\frac{b+3}{2}\right)$
$D\left(a,\frac{b+3}{2}\right)$
$m_{PD}=0$
$\frac{b+3}{2}-1=0$
$b+3-2=0$
$b=-1$
$E\left(\frac{b+a}{2},\frac{5+b}{2}\right)=\left(\frac{af}{2},2\right)$
$m_{CB}\cdot m_{EP}=-1$
$\left(\frac{5-b}{b-a}\right)=\left(\frac{2-1}{a-1}\right)=-1$
$\left(\frac{6}{-1-a}\right)=\left(\frac{2}{a-3}\right)=-1$
$12=(1+a)(a-3)$
$12=a^2-3a+a-3$
$\Rightarrow a^2-2a-15=0$
$(a-5)(a+3)=0$
$a=5\text{ or }a=-3$
$\text{ Given }ab>0$
$a(-1)>0$
$-a>0$
$a<0$
$a=-3\text{ Accept }$
$\text{ AP line A }(-3,3)P(1,1)$
$y-1=\left(\frac{3-1}{-3-1}\right)(x-1)$
$-2y+2=x-1$
$\Rightarrow x+2y=3\text{ Appling }......$(1)
$\text{ Line }BCB(-1,5)$
$\text{ C }(-3,-1)$
$(y-5)=\frac{6}{2}(x+1)$
$y-5=3x+3$
$y=3x+8$....(2)
Solving (1) & (2)
$x+2(3x+8)=3$
$\Rightarrow 7x+16=3$
$7x=-13$
$x=-\frac{13}{7}$
$y=3\left(-\frac{13}{7}\right)+8$
$=\frac{-39+56}{7}$
$y=\frac{17}{7}$
$x+y=\frac{-13+17}{7}=\frac{4}{7}$