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 PBC/ABC = x+y-2/5 ?

1 Answer
Mar 31, 2018

(PBC)/(ABC)=(x_P+y_P-2)/5

Explanation:

A-=(-6,3)

B-=(-3,5)

C-=(4,-2)

P-=(x_P,y_P)

(PBC)/(ABC)=x_P+y_P-2/5

PBC=1/2xxBCxxPD
D is the foot of the perpendicular from P to BC

ABC=1/2xxBCxxAE

E is the foot of the perpendicular from A to BC

(PBC)/(ABC)=(PD)/(AE)

Slope of the straight line passing through B-=(-3,5) and C-=(4,-2) is given by m=((-2-5))/((4-(-3)))=-7/7=-1

Intercept of the straight line passing through B-=(-3,5) and C-=(4,-2) is given by c=5-(-1)xx(-3)=5-3=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+y-2=0

PD=(x_P+y_P-2)/sqrt(1^2+2^2)

AE=(-6+3-2)/sqrt(1^2+2^2)

(PD)/(AE)=(x_P+y_P-2)/(-6+3-2)=(x_P+y_P-2)/5

(PBC)/(ABC)=(x_P+y_P-2)/5