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).
Answer:-
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]
[Pleft( 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]
[Qleft( 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]
[Rleft( x_3 , y_3 right) = left( 1, – 4 right)]
Thus,
[Pleft( x_1 , y_1 right), Qleft( x_2 , y_2 right), Rleft( 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)