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Question

Using co-ordinate geometry, prove that the
three altitudes of any triangle are concurrent.

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Solution

Choose one side BC=2a along xaxis with its mid-point O as origin so that the point B and C are (a,0) and (a,0) respectively. Let the third vertex be the point (h,k).
AD,BE and CF are three altitudes whose equations are
AD x=h or x+0.yh=0
BE y0=(ha)k(x+a)
Or x(ha)+ky+a(ha)=0
CF Replace a by a in (2)
Or x(h+a)+kya(h+a)=0
The three lines will be concurrent if =0
Or ∣ ∣ ∣10hhaka(ha)h+aka(h+a)∣ ∣ ∣
Apply R3R2 and two rows become identical.
=0 concurrent.

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