Problem Solving 81 Page 99 SBT Math 10 – Kite>

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Topic

In the coordinate plane Oxygenfor triangle ABC yes A(-thirty first), REMOVE(3 ; 5), OLD(3 ; -4). Call WOOD, H, I are the centroid, orthocenter, and circumcenter of the triangle, respectively ABC.

a) Make equations for the lines AB, BC, AC

b) Find the coordinates of the points WOOD, H, I

c) Calculate the area of ​​the triangle ABC

Solution method – See details

a) Find the VTPTs of the lines AB, BC, AC then write PTTQ

b) Parameterizing the coordinates of the points WOOD, H, I (necessary)

Step 1: Find the coordinates of the center of gravity WOOD according to the formula \(\left\{ \begin{array}{l}{x_G} = \frac{{{x_A} + {x_B} + {x_C}}}{3}\\{y_G} = \frac{ {{y_A} + {y_B} + {y_C}}}{3}\end{array} \right.\)

Step 2: Solve the system PT: \(\left\{ \begin{array}{l}\overrightarrow {AH} .\overrightarrow {BC} = 0\\\overrightarrow {BH} .\overrightarrow {AC} = 0\ end{array} \right.\) to find the orthocenter H

Step 3: Solve the system PT: \(\left\{ \begin{array}{l}IA = IB\\IA = IC\end{array} \right.\) to find the coordinates of the center I

Step 4: Calculate distance from A arrive BC is the height ofABC

Step 5: Calculate the length BC then calculate the areaABC

Detailed explanation

a) We have: \(\overrightarrow {AB} = (6;6),\overrightarrow {BC} = (0; – 9),\overrightarrow {AC} = (6; – 3)\)

+ Choose \(\overrightarrow {{n_1}} = (1; – 1)\) satisfying \(\overrightarrow {{n_1}} .\overrightarrow {AB} = 0\). Then AB pass A(-3 ; -1) and get \(\overrightarrow {{n_1}} = (1; – 1)\) should have PT:

xy + 2 = 0

+ Choose \(\overrightarrow {{n_2}} = (1;0)\) satisfying \(\overrightarrow {{n_2}} .\overrightarrow {BC} = 0\). Then BC pass REMOVE(3 ; 5) and get \(\overrightarrow {{n_2}} = (1;0)\) should have PT: x – 3 = 0

+ Choose \(\overrightarrow {{n_3}} = (1;2)\) satisfying \(\overrightarrow {{n_3}} .\overrightarrow {AC} = 0\). Then AC pass OLD(3 ; -4) and get \(\overrightarrow {{n_3}} = (1;2)\) should have PT:

x + 2y + 5 = 0

b) We have:

+ G is the center of gravityABC so \( \Rightarrow G(1;0)\)

+ Let \(H({x_H};{y_H})\) be the orthocenterABC . We have: \(\overrightarrow {AH} = ({x_H} + 3;{y_H} + 1),\overrightarrow {BH} = ({x_H} – 3;{y_H} – 5)\)

Then\(\left\{ \begin{array}{l}AH \bot BC\\BH \bot AC\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {AH} .\overrightarrow {BC} = 0\\\overrightarrow {BH} .\overrightarrow {AC} = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{ l} – 9({y_H} + 1) = 0\\6({x_H} – 3) – 3({y_H} – 5)\end{array} \right.\Leftrightarrow \left\{ \begin{array }{l}{y_H} + 1 = 0\\2{x_H} – {y_H} – 1 = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l }{x_H} = 0\\{y_H} = – 1\end{array} \right.\)

\( \Rightarrow H(0; – 1)\)

+ Let \(I({x_I};{y_I})\) be the center of the circumcircle of the triangle ABC

We have: \(\overrightarrow {IA} = {( – 3 – {x_I}; – 1 – {y_I})^2} \Rightarrow IA = \sqrt {{{({x_I} + 3)}^2} + {{({y_I} + 1)}^2}} \Rightarrow I{A^2} = {({x_I} + 3)^2} + {({y_I} + 1)^2}\)

\(\overrightarrow {IB} = {(3 – {x_I};5 – {y_I})^2} \Rightarrow IB = \sqrt {{{({x_I} – 3)}^2} + {{({ y_I} – 5)}^2}} \Rightarrow I{B^2} = {({x_I} – 3)^2} + {({y_I} – 5)^2}\)

\(\overrightarrow {IC} = {(3 – {x_I}; – 4 – {y_I})^2} \Rightarrow IC = \sqrt {{{({x_I} – 3)}^2} + {{( {y_I} + 4)}^2}} \Rightarrow I{C^2} = {({x_I} – 3)^2} + {({y_I} + 4)^2}\)

Then \(\left\{ \begin{array}{l}IA = IB\\IA = IC\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}I{A^ 2} = I{B^2}\\I{A^2} = I{C^2}\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}{({x_I } + 3)^2} + {({y_I} + 1)^2} = {({x_I} – 3)^2} + {({y_I} – 5)^2}\\{({x_I} + 3)^2} + {({y_I} + 1)^2} = {({x_I} – 3)^2} + {({y_I} + 4)^2}\end{array} \right. \)

\( \Leftrightarrow \left\{ \begin{array}{l}12{x_I} + 12{y_I} = 24\\12{x_I} – 6{y_I} = 15\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_I} + {y_I} = 2\\4{x_I} – 2{y_I} = 5\end{array} \right.\)\( \Leftrightarrow \left \{ \begin{array}{l}{x_I} = \frac{3}{2}\\{y_I} = \frac{1}{2}\end{array} \right.\)\( \Rightarrow I\left( {\frac{3}{2};\frac{1}{2}} \right)\)

So \(G(1;0),H(0; – 1),I\left( {\frac{3}{2};\frac{1}{2}} \right)\)

c) We have: \(d(A,BC) = \frac{{\left| { – 3 – 3} \right|}}{1} = 6\)

\(\overrightarrow {BC} = (0; – 9) \Rightarrow BC = 9\)

Triangle area ABC is: \(S = \frac{1}{2}AD.BC = \frac{1}{2}.6.9 = 27\)

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