Question

the orthocentre of the triangle formed by the lines x-2y+9=0,x+y-9=0,2x-y-9=0

Answer 1

Answer on Question #51347 - Math - Analytic Geometry

The orthocentre of the triangle formed by the lines $x - 2y + 9 = 0, x + y - 9 = 0, 2x - y - 9 = 0$ is ?

Solution

Let's find the coordinates of vertices $A, B, C$ of triangle:

$$\left\{ \begin{array}{l} x - 2 y + 9 = 0 \\ x + y - 9 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x - 2 y + 9 = 0 \\ 3 y - 1 8 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x - 1 2 + 9 = 0 \\ y = 6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 3 \\ y = 6 \end{array} \right.. \text{So}, A(3,6)$$ $$\left\{ \begin{array}{l} x - 2 y + 9 = 0 \\ 2 x - y - 9 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x - 2 y + 9 = 0 \\ y = 2 x - 9 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x - 2 (2 x - 9) + 9 = 0 \\ y = 2 x - 9 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} - 3 x + 2 7 = 0 \\ y = 2 x - 9 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 9 \\ y = 9 \end{array} \right.. \text{So}, B(9,9)$$ $$\left\{ \begin{array}{l} x + y - 9 = 0 \\ 2 x - y - 9 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x + y - 9 = 0 \\ 3 x - 1 8 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 6 + y - 9 = 0 \\ x = 6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = 3 \\ x = 6 \end{array} \right.. \text{So}, C(6,3)$$

Assume that $O(x, y)$ is a orthocenter of the triangle, then $AO \perp BC$ and $BO \perp AC$, which give that the dot product of vectors $\overrightarrow{AO}$ and $\overrightarrow{BC}$ is zero, the dot product of vectors $\overrightarrow{BO}$ and $\overrightarrow{AC}$ is zero. From $\overrightarrow{AO} = (x - 3, y - 6)$, $\overrightarrow{BC} = (-3, -6)$, $\overrightarrow{BO} = (x - 9, y - 9)$, $\overrightarrow{AC} = (3, -3)$ we obtain:

$$\left\{ \begin{array}{l} \overrightarrow {A O} \cdot \overrightarrow {B C} = 0 \\ \overrightarrow {B O} \cdot \overrightarrow {A C} = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 9 - 3 x - 6 y + 3 6 = 0 \\ 3 x - 2 7 - 3 y + 2 7 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 9 y = 4 5 \\ x = y \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = 5 \\ x = 5 \end{array} \right.$$

Answer: orthocenter of the triangle is $O(5,5)$.

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