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Solve Lesson 83 Page 99 Math 10 – Kite >

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Topic

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

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: 5x + y – 9 = 0

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

3xy + 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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