The vertices of a triangle ABC are A(6,3),B(3,5)&C(4,2) . If the co-ordinate of P are (x,y) . How to prove that PBCABC=x+y25 ?

1 Answer
Mar 31, 2018

PBCABC=xP+yP25

Explanation:

A(6,3)

B(3,5)

C(4,2)

P(xP,yP)

PBCABC=xP+yP25

PBC=12×BC×PD
D is the foot of the perpendicular from P to BC

ABC=12×BC×AE

E is the foot of the perpendicular from A to BC

PBCABC=PDAE

Slope of the straight line passing through B(3,5) and C(4,2) is given by m=(25)(4(3))=77=1

Intercept of the straight line passing through B(3,5) and C(4,2) is given by c=5(1)×(3)=53=2

Equation of the straight line passing through B(3,5) and C(4,2) is given by

y=mx+c

y=1x+2
Rearranging

y+x=2

Expressing in the standard form,

x+y2=0

PD=xP+yP212+22

AE=6+3212+22

PDAE=xP+yP26+32=xP+yP25

PBCABC=xP+yP25