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 $P B \frac{C}{A} B C = x + y - \frac{2}{5}$ $PBC/ABC = x+y-2/5$ ?

Question

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 $P B \frac{C}{A} B C = x + y - \frac{2}{5}$ $PBC/ABC = x+y-2/5$ ?

1 Answer

Impact of this question

1711 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-31T15:58:26

$\frac{P B C}{A B C} = \frac{x_{P} + y_{P} - 2}{5}$ $(PBC)/(ABC)=(x_P+y_P-2)/5$

Explanation:

$A \equiv (- 6, 3)$ $A-=(-6,3)$

$B \equiv (- 3, 5)$ $B-=(-3,5)$

$C \equiv (4, - 2)$ $C-=(4,-2)$

$P \equiv (x_{P}, y_{P})$ $P-=(x_P,y_P)$

$\frac{P B C}{A B C} = x_{P} + y_{P} - \frac{2}{5}$ $(PBC)/(ABC)=x_P+y_P-2/5$

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

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

E is the foot of the perpendicular from A to BC

$\frac{P B C}{A B C} = \frac{P D}{A E}$ $(PBC)/(ABC)=(PD)/(AE)$

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

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

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

$y = m x + c$ $y=mx+c$

$y = - 1 x + 2$ $y=-1x+2$
Rearranging

$y + x = 2$ $y+x=2$

Expressing in the standard form,

$x + y - 2 = 0$ $x+y-2=0$

$P D = \frac{x_{P} + y_{P} - 2}{\sqrt{1^{2} + 2^{2}}}$ $PD=(x_P+y_P-2)/sqrt(1^2+2^2)$

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

$\frac{P D}{A E} = \frac{x_{P} + y_{P} - 2}{- 6 + 3 - 2} = \frac{x_{P} + y_{P} - 2}{5}$ $(PD)/(AE)=(x_P+y_P-2)/(-6+3-2)=(x_P+y_P-2)/5$

$\frac{P B C}{A B C} = \frac{x_{P} + y_{P} - 2}{5}$ $(PBC)/(ABC)=(x_P+y_P-2)/5$

Science

Math

Humanities

... and beyond

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 $P B \frac{C}{A} B C = x + y - \frac{2}{5}$ $PBC/ABC = x+y-2/5$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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 PBCABC=x+y−25PBC/ABC = x+y-2/5 ?

1 Answer

Explanation:

Related questions

Impact of this question

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 $P B \frac{C}{A} B C = x + y - \frac{2}{5}$ $PBC/ABC = x+y-2/5$ ?