FINDING THE COORDINATES OF POINT OF TRISECTION

Question 1 :

Find the coordinates of the points of trisection of the line segment joining the points A(−5, 6) and B(4,−3).

Solution :

=  (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)

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

A(−5, 6) and B(4,−3)

=  (1(4) + 2(-5))/(1 + 2), (1(-3) + 2(6))/(1 + 2)

=  (4 - 10)/3, (-3 + 12)/3

=  -6/3, 9/3

=  (-2, 3)

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

A(−5, 6) and B(4,−3)

=  (2(4) + 1(-5))/(2 + 1), (2(-3) + 1(6))/(1 + 2)

=  (8 - 5)/3, (-6 + 6)/3

=  3/3, 0/3

=  (1, 0)

Question 2 :

The line segment joining A(6,3) and B(−1, −4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.

Solution :

To find the half of the length of AB, we have to find the midpoint of the line segment AB.

Let C be the point which divides the line segment AB in the ratio 1 : 1 

  =  (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)

  =  (1(-1) + 1(6))/(1 + 1), (1(-4) + 1(3))/(1 + 1)

  =  (-1 + 6)/2, (-4 + 3)/2

  =  (5/2, -1/2)

A is the point which divides the line segment in the ratio 1 : 2 internally.

A(6, 3)  =  (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)

(1(-1) + 2x1)/(1 + 2), (1(-4) + 2y1)/(1 + 2)  =  (6, 3)

(-1 + 2x1)/3, (-4 + 2y1)/3  =  (6, 3)

Equating the x and y -coordinates 

(-1 + 2x1)/3  =  6

-1 + 2x1  =  18

  2x1  =  19

  x1  =  19/2

(-4 + 2y1)/3  =  3

-4 + 2y1  =  9

  2y1  =  9 + 4

  x1  =  13/2

Hence the point D is (19/2, 13/2).

B is the point which divides the line segment in the ratio 2 : 1 internally.

B(-1, -4)  =  (2x2 + 1(6))/(2 + 1), (2y2 + 1(3))/(2 + 1)

 (2x2 + 6)/3, (2y2 + 3)/3  =  B(-1, -4)

Equating the x and y -coordinates 

(2x2 + 6)/3  =  -1

2x2 + 6  =  -3 

2x2 =  -3 - 6

2x2 =  -9

x2 =  -9/2

(2y2 + 3)/3  =  -4

2y2 + 3  =  -12 

2y2 =  -12 - 3

2y2 =  -15

y2 =  -15/2

Hence the point E is (-9/2, -15/2).

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