Question 11
Question: 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]
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.
Question 12
Question: Find the coordinates of the points on the X-axis which are at a distance of 
units from the point (8, -3). [3 Mark]
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).
Question 13
Question: 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]
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
Question 14
Question: 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]
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)
Question 15
Question: 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]
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).
Question 16
Question: 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]
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)
Question 17
Question: If the point (x, y) is equidistant from (-2, 1) and (1, -3), prove that 6x - 8y = 5. [3 Mark]
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.
Question 18
Question: 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]
Answer:
The co-ordinates of P are given by
Question 19
Question: Find the co-ordinates of the points of trisection of the line segment joining the points (3, -3) and (6, 9). [3 Mark]
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).
Question 20
Question: In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7, 6)? [2 Mark]
Answer: 
Let P (5, 4) divides the line segment AB in the ratio 
:1
The point P divides AB in the ratio 3:2.
