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Question (1):
Find the distance between the following pair of points.[2 Mark]
(i) (-5, 3), (3, 1)
(ii) (4, 5), (-3, 2)
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Answer:
(i) Let A (x1, y1) = (-5, 3)
B (x2, y2) = (3, 1)
The distance between the points is
(ii) A = (4, 5) and B = (-3, 2)
Let (x1, y1) = (4,5) and (x2, y2) = (-3, 2)
The distance between the points is
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Question (2):
Find the distance between the following pairs of points.[Each 2 Marks]
(a) (-a, b) and (a, b)
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Answer:
(a) (x1, y1) = (-a, b) (x2, y2) = (a, b)
The distance between the two points is
Distance between the two points is
Distance between the two points is
Distance between the two points is
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Question (3):
Use distance formula, show that the points (8, 7), (6, 4) and (0, -5) are collinear. [3 Mark]
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Answer:
Let A = (8, 7), B = (6, 4), C = (0, -5)
Out of the line segments AB, BC and AC, if the sum of two of them is equal to the third, then A,
B, C are collinear.
 The three given points A, B, C are collinear.
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Question (4):
A point is equidistant from A (-6, 4) and B (2, -8). Find its co-ordinates, if its abscissa
and ordinate are equal. [2 Mark]
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Answer:
Let P be the point which is equidistant from A and B. Since the abscissa and ordinates are equal,
let (x, x) be the co-ordinates of P.
PA = PB (given)
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Question (5):
The distance between A (1, 3) and B (x, 7) is 5. Calculate the possible values of x.
[2 Mark]
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Answer:
Squaring both sides, we get
25 = x2- 2x + 17
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Question (6):
What points on the x-axis are at a distance of 5 units from the point (5, -4)?
[2 Mark]
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Answer:
Note that all the points on x-axis have y-co-ordinate zero.
Let (x, 0) be the point which is at a distance of 5 units from the point (5, -4).
Using distance formula, we have
Squaring both sides, we get
(5 - x)2+ 16 = 25
 The points are (8, 0) and (2, 0).
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Question (7):
What point on y-axis is equidistant from the points (7, 6) and (-3, 4)? [2 Mark]
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Answer:
Note that all the points on y-axis has x co-ordinate 0.
Let P (0, y) be the point on Y axis which is equidistant from A (7,6) and B (-3,4).
PA = PB (given)
 (7 - 0)2 + (6 - y)2
 (- 3 - 0)2 + (4 - y)2
 Squaring both sides and simplifying, we get
49 + 36 + y2- 12y = 9 + 16 - 8y + y2
-12y + 8y = 25 - 49 - 36
-4y = -60
 The point (0, 15) is equidistant from the points (7,6) and (-3,4).
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Question (8):
A is a point on the y-axis whose ordinate is 5 and B is the point (-3, 1). Calculate the
length of AB. [2 Mark]
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Answer:
The co-ordinates of A are (0,5)
B = (-3,-1) (given)
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Question (9):
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
[2 Mark]
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Answer:
The co-ordinates of B are (11, 0).
(on the x-axis, y = 0, abscissa = x-coordinate = 11)
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Question (10):
A point A is at a distance  from the point (4, 3). Find the co-ordinates of point A,
if its ordinate is twice of its abscissa. [3 Mark]
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Answer:
Ordinate is y-co-ordinate and Abscissa is x co-ordinate.
Let the co-ordinates of A be (x, y).
Then, y = 2x (given)
 A is a point which has co-ordinates (x, 2x) and B (4, 3).
 The two possible points are (3, 6) and (1, 2) .
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Question (11):
Find the perimeter of the triangle formed by the points (5, 0), (4, -2) and (2, -1).
[3 Mark]
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Answer:
Let A (5, 0), B (4, -2) and C (2, -1) be the vertices of the triangle.
Perimeter of the triangle = AB + BC + CA
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Question (12):
Show that the points A (8, 3), B (0, 9) and C (14, 11) are the vertices of an isosceles
right-angled triangle. [3 Mark]
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Answer:
Since BC = CA, it is an isosceles triangle.
Since BC is the largest side, check if the square of the largest side is equal to sum of the squares
of the other two sides.
 ABC is a right-angled triangle. (By converse of Pythagoras theorem) |
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Question (13):
Prove that the points A (1, 1), B (-1, -1) and  from an equilateral triangle
and find its area. [3 Mark]
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Answer:
 ABC is an equilateral triangle
Area of an equilateral triangle
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Question (14):
Prove that the quadrilateral formed by the four points A (-1, -6), B (2, -5), C (7, 2)
and D (4, 1) is a parallelogram. [3 Mark]
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Answer:
 AB = DC and BC=AD
Since the opposite sides of the quadrilateral ABCD are equal, ABCD is a parallelogram.
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Question (15):
Prove that the points A (3, -2), B (7, 6), C (-1, 2) and D (-5, -6), in the given order
form a rhombus. Also find its area. [3 Mark]
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Answer:
Since AB = BC = CD = DA, the four given points taken in order form a rhombus.
To find the diagonals:
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Question (16):
Prove that the points (2, -2), (8, 4), (5, 7) and (-1, 1) are the vertices of a rectangle.
[3 Mark]
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Answer:
Let A = (2, -2), B = (8, 4), C = (5, 7) and D = (-1, 1)
Here AB=DC and AD=BC. Opposite sides of the quadrilateral are equal
 ABCD is a
parallelogram.
To prove it is rectangle, we have to prove that the diagonals are equal.
Since AC=BD, the diagonals of the parallelogram are equal, the parallelogram is a rectangle.
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Question (17):
A square has two opposite vertices at (2, 3) and (4, 1). Find the length of the side of
the square. [2 Mark]
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Answer:
Let the two opposite vertices be A(2,3) and C(4,1).
If each side of the square is a, its diagonal =
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Question (18):
Find the co-ordinates of the circumcentre of the triangle ABC, whose vertices A, B and
C are (4, 6), (0, 4) and (6, 2) respectively. [3 Mark]
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Answer:
The circumcentre is the centre of the circumcircle which passes through the vertices A,
B, C of the triangle.
Let the circumcentre be P(x, y), then
PA = PB = PC (circumradius)
PA = PB
x2- 8x + 16 + y2- 12y + 36 = x2 + y2 - 8y + 16
PA = PC
x2- 8x + 16 + y2- 12y + 36 = x2-12x + 36 + y2- 4y + 4
Subtracting (2) from (1), we have
5y = 15
y = 3
Substituting y=3 in (1), we get
2x + 3 = 9
2x = 6
x = 3
 Circumcentre = (3, 3)
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Question (19):
Find the points which are at a distance 5 from (3, 4) and at a distance 13 from (5, 2).
[3 Mark]
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Answer:
Let (x, y) be the required point.
Then, 
(x - 3)2+ (y - 4)2 = 25 (by squaring both sides)
x2- 6x + 9 + y2- 8y + 16 = 25
(x - 5)2 + (y - 12)2 = 169 (squaring)
Subtracting (2) from (1), we have
4x + 16y = 0
Substituting x = -4y in (1), we get
(4y)2+ y2 - 6(-4y) - 8y = 0
If y = 0, x = -4y = -4(0) = 0
The two required points are (0,0), |
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Question (20):
A point P lies on x-axis and another point Q lies on y-axis.[3 Mark]
(i) Write the ordinate of point P.
(ii) Write the abscissa of point Q.
(iii) If the abscissa of point P is -12 and the ordinate of point Q is -16, calculate the length of the
line segment PQ.
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Answer:
(i) The ordinate (y-co-ordinate) is 0 since P lies on x-axis.
(ii) The abscissa of Q is 0, since Q lies on y-axis.
(iii) The co-ordinates of P are (-12,0).
The co-ordinates of Q are (0,-16).
Length of 
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Question (21):
Given A = (3, 1) and B = (0, y-1), find y if AB = 5. [2 Mark]
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Answer:
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Question (22):
Given A = (x+2, -2) and B = (11, 6). Find x if AB = 17. [2 Mark]
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Answer:
(9-x)2 = 289 - 64
(9-x)2 = 225
-x = 6 or -x = -24
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Question (23):
The centre of a circle is (2x-1, 3x+1). Find x if the circle passes through (-3, -1) and
the length of the diameter is 20 units. [3 Mark]
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Answer:
Radius = 10 units
The distance between the centre of the circle and any point on the circle = radius
 (Squaring both sides)
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Question (24):
Find the distance between the following pairs of points. [Each 2 Mark]
(i) (2, 4), (6, 1)
(ii) (9, -13), (2, 11)
(iii) (7, -2), (-8, 6)
(iv) (5, 1), (5, -4)
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Answer:
(i) Distance between two points A and B given by the formula, A(x1, y1), B(x2, y2)
Let A(x1, y1) = (2, 4),
B(x2, y2) = (6, 1)
= 5 units
(ii) Let A(9, -13), B(2, 11)
= 25 units.
(iii) (7, -2), (-8, 6)
Let A(7, -2), B(-8, 6)
= 17 units.
(iv) (5, 1), (5, -4)
Let A(5, 1), B(5, -4)
= 5 units
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Question (25):
Find the measures of the distance between the following: [Each 2 Mark]
(i) (a, 2a), (-2a, 6a)
(iii) (3.2, 8), (-1.6, 1.6)
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Answer:
(i) Let A(a, 2a), B(-2a, 6a)
= 5a units
(ii) 
Let 
= 3 units
(iii) (3.2, 8), (-1.6, 1.6)
Let A(3.2, 8), B(-1.6, 1.6)
AB = 8 units
Distance between the points,
Substituting the values
= 12 units
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