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Question (1):
If the distance between the points (5, -6) and (x, -1) is 13, find the positive values of
x. [3 Mark]
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Answer:
The distance between two points is given by 
Given A (5, -6), B (x, -1), AB = 13
Squaring both the sides,
Solving this quadratic equation,
Given x is a positive value, hence x = 17.
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Question (2):
Show that the point A (7, 7) is equidistant from the points B (3, 4) and C (4, 3).
[3 Mark]
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Answer:
Given, A (7, 7) is equidistant from B (3, 4) and C (4, 3)
Hence AB is given by distance formula, 
= 5 units
= 5 units
Hence AB = AC
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Question (3):
Calculate the distance between the points P(3, 5), Q(6, 7). Leave your answer in root
form. [2 Mark]
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Answer:
 A(x1, y1), B(x2, y2) distance between AB by distance formula is,
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Question (4):
Calculate the distance between the points (-3, 5) and (4, -6), leave your answer in
surd form. [2 Mark]
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Answer:
Let A(-3, 5) and B(4, -6)
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Question (5):
Prove that the points A(-5, 4), B(-1, -2) and C(5, 2) are the vertices of an isosceles
right angled triangle. [3 Mark]
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Answer:
Given that ABC is a triangle.
If ABC is an isosceles D le, then AB = BC.
Hence ABC is an isosceles triangle.
If it is a right angled triangle then,
AC2 = AB2 + BC2
AC2 = 104 units
= 2 x 52
= 104 units
Hence AC2 = AB2 + BC2 and AB = BC
 is an isosceles right angled triangle.
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Question (6):
Prove that the points (3, 1), (4, 3), (0, 3) and (-1, 1) are the vertices of a
parallelogram. Or diagonals bisect each other. [3 Mark]
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| Show that the opposite sides are equal
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Answer:
Let A(3, 1), B(4, 3), C(0, 3), D(-1, 1) the vertices of a (parallelogram) Quadrilateral.
In a parallelogram opposite sides are equal and parallel.
Hence if AB = CD and AD = BC, then ABCD is a parallelogram.
By distance formula,
Hence one pair of opposite sides are equal.
Consider AD and BC,
= 4 units
= 4 units
Hence AB = DC and AD = BC
 ABCD is a parallelogram. |
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Question (7):
Prove that the triangle PQR with vertices P (5, -3), Q (5, 0), R (2, 0) is an isosceles
triangle. [2 Mark]
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Answer:
If triangle PQR is an isosceles triangle then either PQ = QR,
(or any 2 sides are equal)
PQ 
= 3 units
PQ = QR Hence  is isosceles. |
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Question (8):
Find the Co-ordinates of the points whose abscissa is 10 and which are at a distance
of 10 units from (2, -3). [3 Mark]
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Answer:
Let the point A whose abscissa is 10 have the co-ordinates as A (10, y).
Another point B (2, -3)
AB = 10 units (given)
According to distance formula,
Squaring both the sides,
100 = y2 + 6y + 73
Solving the quadratic,
 y2 + 6y - 27 = 0
 y2 + 9y - 3y - 27 = 0
 y(y + 9) - 3(y + 9) = 0
 (y + 9(y - 3) = 0
 y = -9 or y = 3
Vertices of ordinates are -9 or 3
Points are, (10, -9) or (10, 3)
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Question (9):
Prove that A (-4, 3), B (0, 9), C (-3, 11) and D (-7, 5) form a parallelogram ABCD.
[3 Mark]
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Answer:
If ABCD are the vertices of the parallelogram, then their opposite sides are equal in
length and parallel.
 AB = CD and BC = AD
Hence AB = CD
Hence BC = AD
 ABCD is a parallelogram.
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Question (10):
Prove that P (-5, 4), Q (-1, -2), R (5, 2) and S (1, 8) form a square PQRS.
[3 Mark]
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Answer:
According to the property of square, if PQRS is a square then all the sides are equal PR = QS
Hence PQ = QR = RS = SP.
 All the sides are equal.
One of the diagonals,
Hence PR = SQ
 Diagonals are equal.
Hence PQRS is a square. |
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Question (11):
The vertices P(2, -3), Q(5, 1), R(1, 4) and S(-2, 0) form a square. Verify it and show
that each side of the square is 5 units. [3 Mark]
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Answer:
If PQRS is a square, then PQ = QR = RS = SP.
= 5 units
= 5 units
= 5 units
= 5 units
Hence PQRS is a square of 5 units. |
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Question (12):
Find the coordinates of the points on the X-axis which are at a distance of  units
from the point (8, -3). [3 Mark]
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Answer:
Points on the X-axis has y coordinate or ordinate = 0. Let the x coordinate or abscissa
be a.
Let A(a, 0) and B(8, -3)
Squaring on both sides,
Hence the coordinates of the points are (14, 0) or (2, 0).
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Question (13):
Find the coordinate of the point P which is equidistant from A (5, 5), B (-2, 4) and C (-
2, -2). Hence find AP. [3 Mark]
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Answer:
Given P(x, y) is equidistant from A(5, 5), B(-2, 4), C(-2, -2)
Then PA = PB = PC
According to distance formula,
Given PA = PB
Squaring both the sides,
(5 - x)2 + (5 - y)2 = (-2 - x)2 + (4 - y)2
PB = PC
Squaring both the sides,
(-2 - x)2 + (-2 - y)2 = (-2 - x)2 + (4 - y)2
Substituting the values of y = 1 in 15 - 7x - y = 0
Hence the coordinates of P(2, 1)
 PA = 5 units
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Question (14):
Abscissa of a point P equals its ordinate. If P is equidistant from A (5, -3) and B (8, -
4), find the coordinates of the point P. [3 Mark]
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Answer:
Given, Abscissa = ordinate for a point.
P = (a, a)
By distance formula,
But PA = PB
Squaring both the sides,
(a - 5)2 + (a + 3)2 = (a - 8)2 + (a + 4)2
a2 + 25 - 10a + a2 + 6a + 9 = a2 + 64 - 16a + a2 + 16 + 8a
 Coordinates of point P(11.5, 11.5)
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Question (15):
The ordinate of point P is greater than its abscissa by 3 units. Its distance from a point
A (-4, -8) is 17 units. Find the coordinates of the point P if it in the first quadrant.
[3 Mark]
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Answer:
Let the coordinates of point P (x, y)
given, y = x + 3 and PA = 17, Now P(x, x+3)
By distance formula,
Squaring both the sides,
172 = (x + 4)2 + (x + 11)2
289 = (x + 4)2 + (x + 11)2
289 = x2 + 16 + 8x + x2 + 121 + 22x
289 = 2x2 + 30x + 137
0 = 2x2 + 30x + 137 - 289
0 = 2x2 + 30x - 152
0 = x2 + 19x - 4x - 76
0 = x(x + 19) - 4(x + 19)
0 = (x + 19)(x - 4)
But given abscissa and ordinate lie in 1st quadrant.
 Both are +ve values.
Hence x = 4 then y = x + 3 = 4 + 3 = 7
 Coordinates of P(4, 7).
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Question (16):
Abscissa of a point P is twice its ordinate. If the point P is equidistant from (+2, -5) and (-3, 6), find its co-ordinates. [3 Mark]
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Answer:
Given P(x, y) be the point.
Given, abscissa is twice its ordinate.
P is equidistant from A (2, -5), B (-3, 6)
by distance formula,
Given PA = PB
Squaring both the sides,
Hence x = 16
 P (16, 8)
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Question (17):
If the point (x, y) is equidistant from (-2, 1) and (1, -3), prove that 6x - 8y = 5. [3 Mark]
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Answer:
Given P(x, y) is equidistant from A (-2, 1) and B (1, -3)
 PA = PB
By distance formula,
Given, PA = PB
Squaring both sides,
(x + 2)2 + (y - 1)2 = (x - 1)2 + (y + 3)2
Hence proved.
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Question (18):
Calculate the co-ordinates of the point P which divides the line joining A (-3, 3) and B
(2, -7) internally in the ratio 2: 3. [2 Mark]
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Answer:
The co-ordinates of P are given by
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Question (19):
Find the co-ordinates of the points of trisection of the line segment joining the points
(3, -3) and (6, 9). [3 Mark]
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Answer:
AB is divided into 3 equal parts
 P is the point on AB, which divides AB in the ratio 1:2.
Let (x1, y1) be the co-ordinates of P which divides A (3, -3) and B (6, 9) in the ratio 1:2.
Then 
 The co-ordinates P are (4, 1)
 are the co-ordinates of Q, Q divides AB in the ratio 2:1
 The co-ordinates of Q are (5, 5). |
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Question (20):
In what ratio does the point (5, 4) divide the line segment joining the points (2, 1)
and (7, 6)? [2 Mark]
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Answer:
Let P (5, 4) divides the line segment AB in the ratio  :1
 The point P divides AB in the ratio 3:2. |
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Question (21):
Calculate the ratio in which the line segment joining (3, 4) and (-2, 1) is divided by the y-axis. [2 Mark]
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Answer:
On the y-axis, the abscissa is 0. Let the point (0, y) divide the line joining A (3, 4) and (-2, 1) in
the ratio k: 1. Then, using section formula, we have
 The required ratio is k: 1
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Question (22):
In what ratio the line joining the points (4, 2) and (3, -5) is divided by the x-axis?
Also, find the co-ordinates of the point of division. [2 Mark]
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Answer:
On the x-axis, y-co-ordinate is zero.
Let P(x, 0) be the point which divides the line joining (4, 2) and (3, -5) in the ratio k: 1.
Using section formula, we have
The required ratio is k: 1=
= 2:5
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Question (23):
The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such
that . Find the co-ordinates of P [3 Mark]
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Answer:
 P divides the line segment AB in the ratio 4:1.
Let (x, y) be the co-ordinates of P.
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Question (24):
P is a point on the line joining A (4, 3) and B (-2, -6) such that 5AP = 2BP. Find the
co-ordinates of P. [2 Mark]
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Answer:
Let the co-ordinates of P be (x, y).
Using Section Formula, we have
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Question (25):
In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8)? Also, find the
value of a. [2 Mark]
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Answer:
Let P (a,6) divide the line segment A(-4,3) and B(2,8) in the ratio  :1. Then, taking
the y-co-ordinate of P, we have
x-co-ordinate of P is given by
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