Question

find the coordinates of points of trisection of line segment AB with A(2,7) B(-4,-8)​

Answer 1

Answer:

The coordinates of the points of trisection seg AB are (0, 2) and (-2, -3).

Step-by-step explanation:

Let P (x1,y1) and Q (x2,y2) divide the line AB into 3 equal parts.

AP = PQ = QB

AP/PB = AP/(PQ+QB) = AP/(AP+AP) = AP/2AP = 1/2

So, P divides AB in the ratio 1 : 2.

x1 = 1 x (-4) + 2 x 2 / 2 + 1 = 0

y1 = 1 x (-8) + 2 x 7 / 2 + 1 = 2

(x1,y1) = (2,0)

AQ/QB = AP + PB / QB = QB + QB / QB = 1 / 2

x2 = 2 x (-4) + 1 x 2 / 2 + 1 = - 2

y2 = 2 x (-8) + 1 x 7 / 2 + 1 = - 3

(x2,y2) = (-2,-3)

Thus, the coordinates of the point of intersection are

P (x1,y1) = (0,2)

Q (x2,y2) = (-2,-3)