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  • Let the circumcentre of a triangle with vertices <img alt="\mathrm{A(a, 3), B(b, 5) and \: C(a, b), a b>0 \: be P(1,1)}" src="/latex-image/?%5Cmathrm%7BA%28a%2C%203%29%2C%20B%28b%2C%205%29%20and

Let the circumcentre of a triangle with vertices \mathrm{A(a, 3), B(b, 5) and \: C(a, b), a b>0 \: be P(1,1)}. If the line \mathrm{ A P} intersects the line \mathrm{ B C} at the point \mathrm{Q\left(k_{1}, k_{2}\right), then\: k_{1}+k_{2}}  is equal to:

Option: 1

2


Option: 2

\frac{4}{7}


Option: 3

\frac{2}{7}


Option: 4

4


Answers (1)

best_answer

The circumcentre of a triangle is the point of intersection the perpendicular bisector of the sides of a triangle.

\begin{aligned} &\mathrm{(A C)_m \times(S P)_m=-1 }\\ &\mathrm{\frac{1}{0} \times(s p)_m=-1 }\\ &\mathrm{(sp)_m=0 }\\ &\mathrm{\frac{\frac{b+3}{2}-1}{a-1}=0} \\ &\mathrm{b=-1} \\ &\mathrm{(AB)_{m} \times \(RP)_{m}=-1}\\ &\mathrm{\left(\frac{2}{b-a}\right)\left(\frac{3}{\frac{4}{2}-1}\right)=-1\;\;\;\;\;\;\;\left | \because b=-1 \right .} \\ &\mathrm{\frac{6}{(a+1)\left(\frac{a-3}{2}\right)}=1}\\ &\mathrm{a^2-2 a-15=0} \\ &\mathrm{a=-3, \quad or \quad a=5}\\ \end{aligned}

\begin{aligned} &\text{Given }\mathrm{ a b>0}\\ &\mathrm{a(-1)>0} \\ &\mathrm{a<0 }\\ &\mathrm{a=-3} \cdot \text{accept} \end{aligned}

Equation of line AP which is passes through \text{ A(-3,3), and point P(1,1) }\\

\begin{gathered} \mathrm{(y-1)=\left(\frac{3-1}{-3-1}\right)(x-1) }\\ \mathrm{-2 y+2=x-1} \\ \mathrm{x+2 y=3........(1)} \end{gathered} \\

\text{Equation of line BC}\\ \begin{gathered} \mathrm{(y-5)=\frac{6}{2}(x+1) }\\ \mathrm{y=3 x+8}............(2) \end{gathered}

\text{Intersection of line AP and BC is point }\left(K_1, K_2\right) \\

\text{So from } \mathrm{eq^{n}} \text{(1) and (2)}\\ \therefore \text{point Q } \mathrm{\left ( K_{1}, K_{2} \right )\Rightarrow (x, y) \Rightarrow\left(\frac{-13}{7}, \frac{17}{7}\right)} \\ \text{so } \mathrm{ K_{1}+K_{2}=\left(\frac{4}{7}\right)}

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