1
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.
Open in App
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 =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
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
Suggest Corrections
1
Similar questions
