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  • Find the co-ordinates of the orthocenter of the triangle whose vertices are the points <img alt="\mathrm{\left ( 1,-2 \right ),\left ( 3,1 \right )}" src="https://entrancecorner.oncodecogs.com

Find the co-ordinates of the orthocenter of the triangle whose vertices are the points \mathrm{\left ( 1,-2 \right ),\left ( 3,1 \right )} & \mathrm{\left ( -2,3 \right )}.

Option: 1

\mathrm{\left ( \frac{37}{19},\frac{7}{19} \right )}


Option: 2

\mathrm{\left ( \frac{19}{4},\frac{3}{7} \right )}


Option: 3

\mathrm{\left ( \frac{37}{19},\frac{19}{7} \right )}


Option: 4

\mathrm{\left ( \frac{7}{3},\frac{19}{4} \right )}


Answers (1)

best_answer

\mathrm{BC}  is the straight line joining \mathrm{\left ( 3,1 \right )} & \mathrm{\left ( -2,3 \right )} Its siope is \mathrm{\frac{3-1}{-2-3}=-\frac{2}{5}}
If \mathrm{AD} is drawn perpendicular to \mathrm{BC}, Its slope is \mathrm{5/2 }
\mathrm{\therefore } Its equation is \mathrm{y+2=5 / 2(x-1) \text\ { i.e. }\ 5 x-2 y=9 }
If \mathrm{BE } is drawn perpendicular to \mathrm{AC }, We can show that its equation is
\mathrm{y-1=\frac{3}{5}(x-3) \text { i.e. } 3 x-5 y=4 \text { i.e. } 3 x-5 y=4}

The orthocenter is the point of intersection of the altitudes \mathrm{AD} and \mathrm{BE}. Solving their equations we will get co-ordinates of the orthocenter the orthocenter is \mathrm{\left ( \frac{37}{19},\frac{7}{19} \right )}

 

 

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