Toán 10 Viết ptct

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Cựu TMod Toán
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19 Tháng một 2019
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Gọi tọa độ của H bất kì trên AM là (x,y)
Ta có: AB=(51,41)=(4,3)uAB=(4,3)nAB=(3,4)AB:3(x1)4(y1)=03x4y+1=0\overrightarrow{AB}=(5-1,4-1)=(4,3)\Rightarrow \overrightarrow{u}_{AB}=(4,3)\Rightarrow \overrightarrow{n}_{AB}=(3,-4)\Rightarrow AB : 3(x-1)-4(y-1)=0\Leftrightarrow 3x-4y+1=0
AC=(41,141)=(3,13)nAC=(13,3)AC:13(x1)+3(y1)=013x+3y16=0\overrightarrow{AC}=(-4-1,14-1)=(-3,13)\Rightarrow \overrightarrow{n}_{AC}=(13,3)\Rightarrow AC : 13(x-1)+3(y-1)=0\Leftrightarrow 13x+3y-16=0
Lại có: {d(H,AB)=3x4y+132+42=3x4y+15d(H,AC)=13x+3y16132+32=13x+3y16178\left\{\begin{matrix} d_{(H,AB)}=\frac{|3x-4y+1|}{\sqrt{3^2+4^2}}=\frac{|3x-4y+1|}{5}\\ d_{(H,AC)}=\frac{|13x+3y-16|}{\sqrt{13^2+3^2}}=\frac{|13x+3y-16|}{\sqrt{178}} \end{matrix}\right.
d(H,AB)=d(H,AC)d_{(H,AB)}=d_{(H,AC)}\Rightarrow 3x4y+15=13x+3y16178\frac{|3x-4y+1|}{5}=\frac{|13x+3y-16|}{\sqrt{178}}
 
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