Question

Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(–4, –8).

Answer 1

Let P \[\left( x_1 , y_1 \right)\] and Q \[\left( x_2 , y_2 \right)\] divide the line AB into 3 equal parts. 
AP = PQ = QB

\[\frac{AP}{PB} = \frac{AP}{PQ + QB} = \frac{AP}{AP + AP} = \frac{AP}{2AP} = \frac{1}{2}\]

So, P divides AB in the ratio 1 : 2. 

\[x_1 = \frac{1 \times \left( - 4 \right) + 2 \times 2}{2 + 1} = 0\]

\[ y_1 = \frac{1 \times \left( - 8 \right) + 2 \times 7}{2 + 1} = 2\]

\[\left( x_1 , y_1 \right) = \left( 0, 2 \right)\]

Also, \[\frac{AQ}{QB} = \frac{AP + PB}{QB} = \frac{QB + QB}{QB} = \frac{2}{1}\]

\[x_2 = \frac{2 \times \left( - 4 \right) + 1 \times 2}{2 + 1} = - 2\]

\[ y_2 = \frac{2 \times \left( - 8 \right) + 1 \times 7}{2 + 1} = - 3\]

\[\left( x_2 , y_2 \right) = \left( - 2, - 3 \right)\]

Thus, the coordinates of the point of intersection are
P \[\left( x_1 , y_1 \right) = \left( 0, 2 \right)\]

Q \[\left( x_2 , y_2 \right)\] =\[\left( - 2, - 3 \right)\]