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 # ?

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Answer 1 · 2018-03-31T15:58:26

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

#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#