Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(-4, -8).
Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(-4, -8).
AP = PQ = QR = RB
\[\frac{AP}{PB} = \frac{AP}{PQ + QR + RB} = \frac{AP}{AP + AP + AP} = \frac{AP}{3AP} = \frac{1}{3}\]
\[x_1 = \left( \frac{6 \times 1 + 3 \times \left( - 14 \right)}{1 + 3} \right) = \frac{6 - 42}{4} = - 9\]
\[ y_1 = \left( \frac{1 \times \left( - 2 \right) + 3 \times \left( - 10 \right)}{1 + 3} \right) = \frac{- 2 - 30}{4} = - 8\]
\[P\left( x_1 , y_1 \right) = \left( - 9, - 8 \right)\]
Now,
\[\frac{PQ}{QB} = \frac{PQ}{QR + RB} = \frac{PQ}{PQ + PQ} = \frac{1}{2}\]
\[x_2 = \left( \frac{1 \times 6 + 2 \times \left( - 9 \right)}{1 + 2} \right) = - 4\]
\[ y_2 = \left( \frac{1 \times \left( - 2 \right) + 2 \times \left( - 8 \right)}{1 + 2} \right) = - 6\]
\[Q\left( x_2 , y_2 \right) = \left( - 4, - 6 \right)\]
Now R divides QB into 2 equal parts so, using the midpoint formula we have
\[x_3 = \frac{- 4 + 6}{2} = 1\]
\[ y_3 = \frac{- 6 + \left( - 2 \right)}{2} = - 4\]
\[R\left( x_3 , y_3 \right) = \left( 1, - 4 \right)\]
Thus,
\[P\left( x_1 , y_1 \right), Q\left( x_2 , y_2 \right), R\left( x_3 , y_3 \right)\]
\[\left( - 9, - 8 \right), \left( - 4, - 6 \right)\text { and }\left( 1, - 4 \right)\]
Answer:_
To find the coordinates of the points of trisection of the line segment AB, we need to divide the segment into three equal parts.
First, let’s find the midpoint of AB, which is the point that divides the segment into two equal parts:
Midpoint M = ((x-coordinate of A + x-coordinate of B) / 2, (y-coordinate of A + y-coordinate of B) / 2) = ((2 – 4) / 2, (7 – 8) / 2) = (-1, -0.5)
Next, we need to find the points that divide the segment AM and MB into two equal parts. Let’s call these points P and Q, respectively:
Point P = ((2x + x-coordinate of A) / 3, (2y + y-coordinate of A) / 3) Point Q = ((2x + x-coordinate of B) / 3, (2y + y-coordinate of B) / 3)
where (x, y) are the coordinates of the point of trisection.
Substituting the coordinates of A, B, M, and the formulas for P and Q into the above equations, we get:
For point P: (2x + 2)/3 = (2 – 4)/3 (2y + 7)/3 = (-8 + 7)/3
Solving for x and y, we get:
x = (-2 – 2)/2 = -2 y = (7 – 3)/2 = 2
So the coordinates of point P are (-2, 2).
For point Q: (2x – 4)/3 = (-4 – 2)/3 (2y – 8)/3 = (-8 + 7)/3
Solving for x and y, we get:
x = (-6 + 4)/2 = -1 y = (-1 – 1)/2 = -1
So the coordinates of point Q are (-1, -1).
Therefore, the coordinates of the points of trisection of the line segment AB are (-2, 2), (-1, -1), and the midpoint (-1, -0.5).
Chapter 5. Co-ordinate Geometry – Practice Set 5.2 (Page 115)