# 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  ?

Mar 31, 2018

$\frac{P B C}{A B C} = \frac{{x}_{P} + {y}_{P} - 2}{5}$

#### Explanation:

$A \equiv \left(- 6 , 3\right)$

$B \equiv \left(- 3 , 5\right)$

$C \equiv \left(4 , - 2\right)$

$P \equiv \left({x}_{P} , {y}_{P}\right)$

$\frac{P B C}{A B C} = {x}_{P} + {y}_{P} - \frac{2}{5}$

$P B C = \frac{1}{2} \times B C \times P D$
D is the foot of the perpendicular from P to BC

$A B C = \frac{1}{2} \times B C \times A E$

E is the foot of the perpendicular from A to BC

$\frac{P B C}{A B C} = \frac{P D}{A E}$

Slope of the straight line passing through $B \equiv \left(- 3 , 5\right)$ and $C \equiv \left(4 , - 2\right)$ is given by $m = \frac{\left(- 2 - 5\right)}{\left(4 - \left(- 3\right)\right)} = - \frac{7}{7} = - 1$

Intercept of the straight line passing through $B \equiv \left(- 3 , 5\right)$ and $C \equiv \left(4 , - 2\right)$ is given by $c = 5 - \left(- 1\right) \times \left(- 3\right) = 5 - 3 = 2$

Equation of the straight line passing through $B \equiv \left(- 3 , 5\right)$ and $C \equiv \left(4 , - 2\right)$ is given by

$y = m x + c$

$y = - 1 x + 2$
Rearranging

$y + x = 2$

Expressing in the standard form,

$x + y - 2 = 0$

$P D = \frac{{x}_{P} + {y}_{P} - 2}{\sqrt{{1}^{2} + {2}^{2}}}$

$A E = \frac{- 6 + 3 - 2}{\sqrt{{1}^{2} + {2}^{2}}}$

$\frac{P D}{A E} = \frac{{x}_{P} + {y}_{P} - 2}{- 6 + 3 - 2} = \frac{{x}_{P} + {y}_{P} - 2}{5}$

$\frac{P B C}{A B C} = \frac{{x}_{P} + {y}_{P} - 2}{5}$