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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