Pedagogy of Mathematics (Part II) - Coordinate Geometry, Coordinate Geometry, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, the mid point of a line segment,

1. PEDAGOGY OF
MATHEMATICS –
PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in

2. STD IX - EX 5.3

16. (i) (−2, 3) and (−6,−5)
Solution :
(x1, y1) ==> (-2, 3)
(x2 , y2) ==> (-6, -5)
Midpoint = (-2 - 6)/2, (3 + (-5))/2
= -8/2, -2/2
= (-4, -1)

17. (ii) (8,−2) and (−8,0)
Solution :
(x1, y1) ==> (8, -2)
(x2 , y2) ==> (-8, 0)
Midpoint = (8 - 8)/2, (-2 + 0)/2
= 0/2, -2/2
= (0, -1)

18. (iii) (a, b) and (a + 2b, 2a - b)
Solution :
(x1, y1) ==> (a, b)
(x2 , y2) ==> (a + 2b, 2a - b)
Midpoint = (a + a + 2b)/2, (b + 2a - b)/2
= 2(a + b)/2, 2a/2
= (a + b, 1)

19. (iv) (1/2, -3/7) and (3/2, -11/7)
Solution :
(x1, y1) ==> (1/2, -3/7)
(x2 , y2) ==> (3/2, -11/7)
Midpoint = ((1/2) + (3/2))/2, ((-3/7) + (-11/7))/2
= ((4/2)/2, ((-14/7)/2)
= (1, -1)

20. Solution :
Midpoint of the diameter = Center of the circle
Let the other endpoint be (a, b)
Midpoint of (-3, 7) and (a, b) is (-4, 2).
(-3 + a)/2, (7 + b)/2 = (-4, 2)
By equating the x and y coordinates, we get
(-3 + a)/2 = -4
-3 + a = -8
a = -8 + 3
a = -5 (7 + b)/2 = 2
7 + b = 4
b = 4 - 7
b = -3
Hence the other end is (-5, -3).

21. Solution :
Midpoint of the line segment joining the points (3, 4)
and (p, 7)
(3 + p)/2 , (7 + 4)/2 = (x, y)
(3 + p)/2 , 11/2 = (x, y)
x = (3 + p)/2 and y = 11/2
Since the midpoint lies on the line 2x + 2y + 1 = 0
2(3 + p)/2 + 2(11/2) + 1 = 0
3 + p + 11 + 1 = 0
p + 15 = 0
p = -15

22. Solution :
Let (x1, y1) ==> (2, 4)
(x2, y2) ==> (-2, 3) and (x3, y3) ==> (5, 2)
By using the formula given above,
A (x1 + x3 - x2, y1 + y3 - y2) ==> [(2+5+2), (4+2-3)]
(9, 3)
B (x1 + x2 - x3, y1 + y2 - y3) ==> [(2-2-5), (4+3-2)]
(-5, 5)
C (x2 + x3 - x1, y2 + y3 - y1) ==> [(-2+5-2), (3+2-4)]
(1, 1)
Hence the required vertices are (9, 3) (-5, 5) and (1, 1).

23. Solution :
Midpoint of the chord AB = D
(x1 + x2)/2, (y1 + y2)/2
= (8 + 10)/2, (6 + 0)/2
= 18/2, 6/2
= (9, 3)
Now we we have to find the midpoint of OD, that is E
O(0, 0) D(9, 3)
= (0 + 9)/2, (0 + 3)/2
= (9/2, 3/2)

24. Since it forms a square,
Midpoint of the diagonal AC and BD are equal.
Midpoint of AC :
A (-5, 4) C (5, 2)
= (-5 + 5)/2, (4 + 2)/2
= 0/2, 6/2
= (0, 3)

25. Midpoint of BD :
B (-1, -2) D (a, b)
= (-1 + a)/2, (-2 + b)/2
By equating the x and y coordinates, we get
Midpoint of BD :
B (-1, -2) D (a, b)
= (-1 + a)/2, (-2 + b)/2
By equating the x and y coordinates, we get
(-1 + a)/2 = 0
-1 + a = 0
a = 1
(-2 + b)/2 = 3
-2 + b = 6
b = 6 + 2
b = 8
Hence the required vertex is (1, 8).

26. In order to prove ABCD is a parallelogram, we have to find the point D.
= D (-3+1-0, 6+9-7)
= D (-2, 8)
In ABCD is a parallelogram, then midpoint of diagonals AC and BD will be equal.
Midpoint of AC = (-3 + 1)/2, (6 + 9)/2
= -2/2, 15/2
= (-1, 15/2)

27. Midpoint of BD
B(0, 7) and D(-2, 8)
= (0 - 2)/2, (7 + 8)/2
= -2/2, 15/2
= (-1, 15/2)
Hence ABCD is a parallelogram.

28. Mid point of BC = (3 + (-3))/2, (2 + (-2))/2
= D (0, 0)
Distance between AD :
A(-3, 2) D(0, 0)
= √(x2 - x1)2 + (y2 - y1)2
= √(-3-0)2 + (2-0)2
= √9 + 4
= √13

29. C(-3, -2) D(0, 0)
= √(x2 - x1)2 + (y2 - y1)2
= √(-3-0)2 + (-2-0)2
= √9 + 4
= √13
B(3, 2) D(0, 0)
= √(x2 - x1)2 + (y2 - y1)2
= √(0-3)2 + (0-2)2
= √9 + 4
= √13
Hence the midpoint of the hypotenuse is equidistant
from the vertices.