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**Topic**

In the coordinate plane *Oxygen*for triangle *ABC* yes *A*(−1 ; −2), median line drawn from *REMOVE* and the high line drawn from *OLD* whose equation is 5 . respectively*x* + *y* – 9 = 0 and *x* + 3*y* − 5 = 0. Find coordinates of two points *REMOVE* and *OLD*.

**Solution method – See details**

Step 1: Write the equation of the straight line AB (where VTPT is VTCP of *ONLY*)

Step 2: Solve the system 2 PT *BM* and *AB* to find point coordinates *REMOVE*

Step 3: Parameterize the point *USA* according to PT *BM* and coordinate representation *OLD* according to that parameter

Step 4: Replace the parameter coordinates of the point *OLD* go to PT CH and find the coordinates of the point *OLD*

**Detailed explanation**

Call *BM* is the median line drawn from *REMOVE* \( \Rightarrow BM\) has PT: 5*x* + *y* – 9 = 0

Call *ONLY *is the height line drawn from *OLD* \( \Rightarrow CH\) with PT: *x* + 3*y* − 5 = 0

CH has VTPT \(\overrightarrow {{n_1}} = (1;3)\) \( \Rightarrow CH\) has VTCP \(\overrightarrow {{u_1}} = (3; – 1)\)

We have: \(CH \bot AB\) \( \Rightarrow AB\) goes through *A*(−1 ; −2) and get \(\overrightarrow {{u_1}} = (3; – 1)\) as VTPT, so there is PT:

3*x* – *y* + 1 = 0

Do *REMOVE* is the intersection of *BM* and *AB *so point coordinates *REMOVE* is the solution of the system PT:

\(\left\{ \begin{array}{l}5x + y – 9 = 0\\3x – y + 1 = 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{ l}x = 1\\y = 4\end{array} \right \Rightarrow B(1;4)\)

Since \(M \in BM\) so \(M(t;9 – 5t)\)

According to the assumption, *USA* is the midpoint *AC *\( \Rightarrow C(2t + 1; – 10t + 20)\)

Since \(C \in CH\) so \(2t + 1 + 3( – 10t + 20) – 5 = 0 \Leftrightarrow – 28t + 56 = 0 \Leftrightarrow t = 2\) \( \Leftrightarrow C(5; 0)\)

So \(B(1;4)\) and \(C(5;0)\)

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