Question

Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

Answer 1

Answer:

Let P and Q divides the line segment AB into three equal parts.

i.e. AP = PQ = QB

P divides the line segment AB in the ratio 1 : 2

m1 = 1 , m2 = 2

Coordinates of P

= ( m1x2 + m2x1 /m1 + m2 ,

m1y2 + m2y1 / m1 + m2 )

= (1×-7 + 2×2 / 1+2 ,

1×4 + 2×-2 / 1+2)

=( -7+4 / 3 , 4-4 / 3)

= ( -1 , 0)

Now ,

Q divides the line segment AB in the ratio 2 : 1

m1 = 2 , m2 = 1

Coordinates of Q

= (m1x2 + m2x1 / m1 + m2 ,

m1y2 + m2y1 / m1 + m2)

= (2×-7 + 1×2 / 2+1 ,

2×4 + 1×-2 / 2+1)

= (-12/3 , 6/3)

= (-4 , 2)

Therefore,

Coordinates of P = (-1 , 0)

Coordinates of Q = (-4 , 2)

HOPE THIS WILL HELP YOU ; )