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 =y2−y3x2−x3
⇒ Slope of straight line perpendicular to
QR=−x2−x3y2−y3
Equation of straight line passing through A(a1,b1,) and perpendicular to QR is
y−b1=−x2−x3y2−y3(x−a1) ⇒(x2−x3)x+(y2−y3)y−a1(x2−x3)−b1(y2−y3)=0 ....(1)
Similarly equation of straight line from B and perpendicular to RP is
(x3−x1)x+(y3−y1)y−a2(x3−x1)−b2(y3−y1)=0 ....(2)
and eqn of straight line from C and perpendicular to PQ is
(x1−x2)x+(y1−y2)y−a3(x1−x2)−b3(y1−y2)=0 ....(3)
As staight lines (1),(2) and (3) are given to be concurrent, we should have
∣∣
∣
∣∣x2−x3y2−y3a1(x2−x3)+b1(y2−y3)x3−x1y3−y1a2(x3−x1)+b2(y3−y1)x1−x2y1−y2a3(x1−x2)+b3(y1−y2)∣∣
∣
∣∣=0 ....(4)
Operating R1→R1+R2+R3, we get
∣∣
∣
∣∣00Sx3−x1y3+y1a2(x3−x1)+b2(y3−y1)x1−x2y1+y2a3(x1−x2)+b3(y1−y2)∣∣
∣
∣∣=0
Where
[S=a1(x2−x3)+b1(y2−y3)+a2(x3−x1)+b2(y3−y1)+a3(x1−x2)+b3(y1−y2)]
Expanding along R1
⇒[(x3−x1)(y1−y2)−(x1−x2)(y3−y1)]S=0
⇒[y1−y2x1−x2−y3−y1x3−x1]S=0⇒[mPQ−mPR]S=0⇒S=0[mPQ=mPR⇒PQ∥PR ;which is not possible in ΔPQR]⇒a1(x2−x3)+b1(y2−y3)+a2(x3−x1)+b2(y3−y1) +a3(x1−x2)+b3(y1−y2)=0....(5)⇒x1(a3−a2)+y1(b3−b2)+x2(a1−a3)+y2(b1−b3) +x3(a2−a1)+y3(b2−b1)=0....(6)
(Rearranging the equation (5))
But above condition shows
∣∣
∣
∣∣a3−a2b3−b2x1(a3−a2)+y1(b3−b2)a1−a3b1−b3x2(a1−a3)+y2(b1−b3)a2−a1b2−b1x3(a2−a1)+y3(b2−b1)∣∣
∣
∣∣=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