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Question (1):
In a parallelogram ABCD, the Co-ordinates of the vertices are A(-8, -4), B(6, -4), C(x,
y), D(-3, 3). Find the Coordinates of the point C by applying mid point formula. [3 Mark]
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Answer:
The diagonals in a parallelogram bisect each other. Let the point of bisection be O.
Then by mid point formula,
 Coordinates of C(11, 3). |
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Question (2):
Points A and B have vertices (7, -2) and (a, b) respectively. The Coordinates of the
mid point are (4, -3). Find the values of a and b. [2 Mark]
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Answer:
By mid point formula, Coordinates of mid point can be calculated by,
Hence a = 1, b = -4
 Coordinates of B(1, -4). |
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Question (3):
Find the Coordinates of the point A and B where the line 3x - 2y = 24 cuts the x-axis
and Y-axis respectively. Also find the Coordinates of the mid point of AB. [3 Mark]
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Answer:
Point A(x, 0) lies on the line 3x - 2y = 24
Hence values of x and y should satisfy the equation,
 3x - 2(0) = 24
 3x = 24
 Coordinates of A (8, 0)
Pt B(0, y) lies on the line 3x - 2y = 24
3(0) - 2y = 24
Co-ordinates of B(0, -12)
Mid point of A(8, 0) , B(0, -12) is given by mid point formula,
= (4, -6)
Hence Co-ordinates of mid points are given by (4, -6). |
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Question (4):
Show that the points A(-2, 3), B(4, 0) and C(1, -3) are the vertices of an isosceles
triangle. Find the Co-ordinates of D so that ABCD is a Rhombus. [3 Mark]
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Answer:
If ABC is an isosceles triangle then AC = AB.
 By distance formula
By distance formula
 is an isosceles triangle.
If ABCD is a Rhombus, then diagonals bisect each other.
 .
O is also the mid point of AD
 Coordinates of D (7, -6) |
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Question (5):
A (-4, -2), B(2, 0), C(8, 6) and D(x, y) are the coordinates of the vertices of a
parallelogram. Find the value of x and y. [3 Mark]
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Answer:
The diagonals bisect each other in a parallelogram.
Hence O is the n\mid point of AC and BD.
= (2, 2)
 Coordinates of D(2, 4). |
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Question (6):
A is a point on the positive side of the X-axis and B is a point on the +ve side of Y-
axis. P(4, 5) is the mid point of AB. Find the Co-ordinates of A and B. [3 Mark]
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Answer:
Given A is a point on +ve side of X-axis
 its Coordinates are (x, 0)
B is a point on +ve side of Y-axis
 its Co-ordinates are (0, y).
By mid point formula, then mid points
But given the coordinates of mid point = P (4, 5)
Hence the Coordinates of A(8, 0), B(0, 10).
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Question (7):
The mid point of the joining (a, 2) and (3, 6) is (2, b). Find the numerical values of a,
b. [2 Mark]
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Answer:
By mid point formula,
 The numerical value of a = 1 and b = 4.
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Question (8):
The line segment joining A(2, 3) and (6, -5) is intercepted by x -axis at a point K.
Write down the ordinate of the point K. Hence find the ratio in which K divides AB.
[3 Mark]
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Answer:
The ordinate of point K is 0 as it lies on x-axis by the interception by x-axis of line AB.
Let section formula the Coordinates of K is given by
Hence m:n = 3:5
K divides AB in the ratio 3:5. |
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Question (9):
Prove that the points A(-5, 4), B(-1, -2) and C(5, 2) are the vertices of an isosceles
right angled triangle. Find the Coordinates of D so that ABCD is a square. [3 Mark]
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Answer:
If ABC is a right angled triangle, by Pythagoras theorem,
AB2 + BC2 = AC2
and it is given ABC is also isosceles,
 AB = BC, hence,
AB2 + AB2 = AC2
2AB2 = AC2
AB2 = 16 + 36 = 52 sq.units
|||ly
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AC2 = 102 + 22 = 104 sq. units
AB2 + BC2 = 52 + 52 = 104 sq. units
AC2 = 104 sq. units
Hence AB2 + BC2 = AC2
If ABCD is a square, then diagonals bisect each other at right angles.
 O is the mid point of AC
Also O is the mid point of BD
= (0, 3)
Let Coordinates of D (x, y) then, by mid point formulae,
 Co-ordinates of D (1, 8). |
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Question (10):
ABC is a triangle with vertices A(a, 5), B(6, -2) and C(-7, b). If (4, 2) are the
Coordinates of its centroid, find a and b. [3 Mark]
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Answer:
Centroid is a point where all the median intersect.
Median is line joining the vertex to the mid point of opposite side. Hence by mid point formula,
mid point of AC
and centroid divides the median in the ratio 2:1,
Consider then median from vertex B, By section formula,
Let m = 2, n = 1
Where m = 2, n = 1,
12 = 2x + 6, 6 = 2y - 2
6 = 2x, 8 = 2y
3 = x, 4 = y
Substituting the values of x, y in Co-ordinates of mid point of AC,
Hence a = 13, b = 3 |
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Question (11):
In what ratio does the y-axis divide the line AB, Where A(-4, 1) and B(17, 10).
[2 Mark]
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Answer:
Let P divide the (1995 - year) given line in the ratio m:n.
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Question (12):
Find the co-ordinates of the point which divides the line segment joining the given
points in the given ratio internally (-4, 1) and (17, 10); 1:2. [2 Mark]
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Answer:
Let A(-4, 1), B(17, 10) and ratio m:n = 1:2
Co-ordinates of point dividing the line segment,
= (3, 4)
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Question (13):
The mid point of the line segment AB shown in the diagram, is (4, -3). Write down the
co-ordinates of A and B. [2 Mark]
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Answer:
The co-ordinates of A (x1, 0), B(0, y2)
(points are A on x-axis, B on y-axis)
Substituting,
Hence co-ordinates of A(8, 0), B(0, -6). |
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Question (14):
Calculate the distance between A(7, 3) and B on the x-axis whose abscissa is 11.
[2 Mark]
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Answer:
Point on x-axis. Hence its ordinate = 0.
 Co-ordinates of B(11, 0), A(7, 3).
By distance formula, distance between AB,
= 5 units
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Question (15):
The centre O, of a circle has the co-ordinates (4, 5) and one point on the
circumference is (8, 10), find the co-ordinates of the other end of the diameter of the circle
through this point. [2 Mark]
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Answer:
Let AB be the diameter.
O centre lies on diameter.
A(8, 10), O(4, 5), B(a, b)
O is the centre  mid point of diameter.
10 = 10 + b
Hence co-ordinates of B(0, 0) [B the other end of the diameter] |
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Question (16):
Calculate the ratio in which the line joining A(6, 5) and B(4, -3) is divided by the line y
= 2. [3 Mark]
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Answer:
Let the line y = 2(parallel to x-axis) divide line AB at O.
 Coordinates of O(x, 2) in the ratio m: n.
Then according to section formula,
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Question (17):
Find the area of a triangle whose vertices are (-5, -1), (3, -5) and (5, 2).
[2 Mark]
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Answer:
Area of triangle 
 = 26
= 26 Sq. Units.
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Question (18):
Find the area of the triangle formed by joining the mid points of the sides of the
triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the
given triangle. [3 Mark]
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Answer:
E is the mid point of AB.  the coordinates of E are =  = (1, 0)
D is the mid point of BC.  the coordinates of D are =  = (1, 2)
F is the mid point of CA.  the coordinates of F are =  = (0, 1)
Area of  DEF =
= 
Area of  ABC = 
= 4 Sq. units
Ratio of the area of  DEF to the area of  ABC is .
Area of  DEF = 1 Sq. units, Area of  ABC is 4 Sq. units. Ratio is  . |
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Question (19):
Find the area of the quadrilateral whose vertices, taken in order are (-4, -2), (-3, -5),
(3, -2) and (2, 3). [3 Mark]
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Answer:
Given A (-4, -2), B (-3, -5), C (3, -2), D (2, 3) be the 4 vertices of a quadrilateral. Join BD.
Area of quadrilateral ABCD is equal to sum of the area of the triangle ABD and DBC.
A (-4, -2), B (-3, -5),D (2, 3)
Area of  ABD = 
=  Sq. units.
D (2, 3), B (-3, -5), C (3, -2)
Area of  DBC = 
=  Sq. units
 Area of the quadrilateral ABCD =  = 28 Squints |
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Question (20):
Find the area of rhombus if the vertices are (3, 0), (4, 5) (-1, 4) and (-2, -1) taken in
order. [3 Mark]
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Answer:
Area of rhombus ABCD = area of  ABC + area of  ACD
Area of  ABC = 
= 
Area of  ACD = 
= 
 Area of rhombus ABCD = 12 + 12 =24 Sq.units. |
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Question (21):
A median of a triangle divides it into two triangles of equal areas. Verify this result for
 ABC. Whose vertices are A (4, -6), B (3, -2) and C (5, 2). [3 Mark]
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Answer:
ABC is a triangle, AD is a median. i.e. D is the mid point BC. D (4, 0)
We have to verify that,
Area of  ABD = area of  ADC.
Area of  ABD = 
= 3 Sq. units
Area of  ADC = 
= 3 Sq. units
Thus it is verified thatarea of  ABD = area of  ADC |
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Question (22):
Find the length of the altitude of the triangle, whose vertices are (5, 1), (2, 4) and (-
1, -1). [3 Mark]
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Answer:
Area of  ABC = 
= 12 Sq. units
Area of  ABC =  x base x height = 12 Sq. units
Length of AB =  units
Length of BC =  units
Length of CA =  units
 Altitude AD =  units
Altitude BE =  units
Altitude CF =  units. |
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Question (23):
For what value of 'x' will the points (x, 3), (-5, 6) and (-8, 8) be collinear?
[2 Mark]
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Answer:
Let the points be A(x, 3), B (-5, 6) and C (-8, 8).
To prove A, B, C are collinear, we have to equate the area of  ABC to zero.
Area of  ABC =  = 0
=  = 0
=  = 0  2x + 1 = 0  x = .
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Question (24):
The co-ordinates of A, B and C are (6, 3), (-3, 5) and (4, -2) respectively and P is any
point (x, y). Show that the ratio of the areas of triangles PBC and ABC is . [2 Mark]
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Answer:
= 
=  |
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Question (25):
For what value of 'x', the area of the triangle formed by the points (5, -1), (x, 4) and
(6, 3) is 5.5 Sq. units. [2 Mark]
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Answer:
Let A (5, -1), B (x, 4) and C (6, 3) be the vertices of  ABC.
Area of  ABC is equal to 5.5 or  Sq. units.
4x -25 = 11
4x = 36
x = 9.
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