If $(\alpha, \beta)$ is the orthocentre of the triangle with the vertices $(2,2),(5,1),(4,4)$, then…
- 6
- 5
- $\frac{5}{2}$
- $\frac{7}{2}$
Solution
$\mathrm{A}(2,2), \mathrm{B}(5,1), \mathrm{C}(4,4)$
Slope of $A B=\frac{1-2}{5-2}=\frac{-1}{3}$ $\therefore$ Equation of CG $\begin{aligned} & y-4=3(x-4) \\ & \Rightarrow 3 x-y=8 \end{aligned}$
Slope of $A C=\frac{4-2}{4-2}=1$ $\therefore$ Equation of BF $y-1=-1(x-5) \Rightarrow x+y=6 \ldots \text { (ii) }$
Solving (i) and (ii) we get $(x, y)=(\alpha, \beta)=\left(\frac{7}{2}, \frac{5}{2}\right) \quad \therefore \alpha+\beta=\frac{12}{2}=6$
Asked in: AP EAMCET 2024 (21 May Shift 1)