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Question

Let ABC and PQR be any two triangles in the same plane. Assume that the perpendiculars from the points A,B,C to the sides QR,RP,PQ respectively are concurrent. Using vector methods or otherwise, prove that the perpendiculars from P,Q,R to BC,CA,AB respectively are also concurrent.

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Solution

Let the co-ordinates of the vertices of the ΔABC be A(a1,b1),B(a2,b2) and C(a3,b3) and co-ordinates of the vertices of the ΔPQR be
P(x1,y1),B(x2,y2) and R(x3.y3)
Slope of QR =y2y3x2x3
Slope of straight line perpendicular to
QR=x2x3y2y3
Equation of straight line passing through A(a1,b1,) and perpendicular to QR is
yb1=x2x3y2y3(xa1) (x2x3)x+(y2y3)ya1(x2x3)b1(y2y3)=0 ....(1)
Similarly equation of straight line from B and perpendicular to RP is
(x3x1)x+(y3y1)ya2(x3x1)b2(y3y1)=0 ....(2)
and eqn of straight line from C and perpendicular to PQ is
(x1x2)x+(y1y2)ya3(x1x2)b3(y1y2)=0 ....(3)
As staight lines (1),(2) and (3) are given to be concurrent, we should have
∣ ∣ ∣x2x3y2y3a1(x2x3)+b1(y2y3)x3x1y3y1a2(x3x1)+b2(y3y1)x1x2y1y2a3(x1x2)+b3(y1y2)∣ ∣ ∣=0 ....(4)
Operating R1R1+R2+R3, we get
∣ ∣ ∣00Sx3x1y3+y1a2(x3x1)+b2(y3y1)x1x2y1+y2a3(x1x2)+b3(y1y2)∣ ∣ ∣=0
Where
[S=a1(x2x3)+b1(y2y3)+a2(x3x1)+b2(y3y1)+a3(x1x2)+b3(y1y2)]
Expanding along R1
[(x3x1)(y1y2)(x1x2)(y3y1)]S=0
[y1y2x1x2y3y1x3x1]S=0[mPQmPR]S=0S=0[mPQ=mPRPQPR ;which is not possible in ΔPQR]a1(x2x3)+b1(y2y3)+a2(x3x1)+b2(y3y1) +a3(x1x2)+b3(y1y2)=0....(5)x1(a3a2)+y1(b3b2)+x2(a1a3)+y2(b1b3) +x3(a2a1)+y3(b2b1)=0....(6)
(Rearranging the equation (5))
But above condition shows
∣ ∣ ∣a3a2b3b2x1(a3a2)+y1(b3b2)a1a3b1b3x2(a1a3)+y2(b1b3)a2a1b2b1x3(a2a1)+y3(b2b1)∣ ∣ ∣=0 ...(7)
[Using the fact that as (4)(5) in the same way (6)(7)]
Clearly equaton (7) shows that lines through P and perpendicular to BC, through Q and perpendicular to AB are conurrent.
HENCE PROVED

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