Question

If A(4,-8) , B(3,6) and C(5,-4) are the vertices of a triangle ABC, D is the mid-point of BC and P is a point on AD joined such that AP/PD = 2, find the coordinates of P.

Answer 1

D will be the midpoint of BC of △ABC with coordinates A(4,-8), B(3,6) and C(5,-4)

Then the coordination of D are

(3+5/2, 6-4/2)

that is (4,1)

Now

PD/AP = 2/1

The coordinates of P are

(2x4+1x4/2+1 , 2x-8 + 1x1/2+1)

= (8+4/3, -16 + 1/3) = (12/3, -15/3)

= (4,5)

Hence the coordinates of P are (4, -5)

Answer 2

Answer: The coordinates of P are (4,-2)

Step-by-step explanation:

Here, A(4,-8) , B(3,6) and C(5,-4)

Also, D is the midpoint of the segment BC,

Hence the coordinates of D are,

$(\frac{3+5}{2},\frac{6-4}{2})$

$=(\frac{8}{2},\frac{2}{2})$

$=(4,1)$

Now, according to the question,

P divides the line segment AD in the ratio of 2 : 1,

Thus, by the section formula,

The coordinates of D are,

$(\frac{2\times 4+1\times 4}{2+1},\frac{2\times 1+1\times -8}{2+1})$

$=(\frac{8+4}{3},\frac{2-8}{3})$

$=(\frac{12}{3},\frac{-6}{3})$

$=(4,-2)$