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

Question

1710 views around the world

You can reuse this answer
Creative Commons License

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,

$x+y-2=0$

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

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

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

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

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