Question 11
Question: Find the perimeter of the triangle formed by the points (5, 0), (4, -2) and (2, -1). [3 Mark]
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
Question 12
Question: 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]
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)
Question 13
Question: Prove that the points A (1, 1), B (-1, -1) and 
from an equilateral triangle and find its area. [3 Mark]
Answer: 
ABC is an equilateral triangle Area of an equilateral triangle
Question 14
Question: 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]
Answer: 
AB = DC and BC=AD
Since the opposite sides of the quadrilateral ABCD are equal, ABCD is a parallelogram.
Question 15
Question: 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]
Answer: 
Since AB = BC = CD = DA, the four given points taken in order form a rhombus.
To find the diagonals:
Question 16
Question: Prove that the points (2, -2), (8, 4), (5, 7) and (-1, 1) are the vertices of a rectangle. [3 Mark]
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.
Question 17
Question: A square has two opposite vertices at (2, 3) and (4, 1). Find the length of the side of the square. [2 Mark]
Answer: 
Let the two opposite vertices be A(2,3) and C(4,1).
If each side of the square is a, its diagonal =
Question 18
Question: 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]
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)
Question 19
Question: Find the points which are at a distance 5 from (3, 4) and at a distance 13 from (5, 2). [3 Mark]
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),
Question 20
Question: 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.
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 
