Coordinate geometry 1 √( ) ( )

This is the area of mathematic where Example 1

geometrical relationships are described
(a) Find the between the points
algebraically by reference to the coordinates
(i) A(1, 3) and B(7, 11)
The length of the line segment Solution
Using √( ) ( )
Given two points A(x1, y1) and B(x2, y2) in x-y
plane, the distance between A and B, denoted √( ) ( )
by is √( ) ( ) √
√ = 10
Proof (ii) P(-1, 2) and Q(3, 7)
Solution
Geometrical approach
√( ) ( )
√( ) ( )
√ =5
(b) The points A, B and C have coordinates
A(-3, 2), B(-1, -2) and C(0, n) where n is a
constant. Given that , find the
possible values of n.
Solution
Using Pythagoras theorem Using √( ) ( )
√( ( )) ( ( ))
√ ( )
=√
( ) ( )
√( ( )) ( )
√( ) ( ) √ ( )
Vector method approach =√
Bu t
AB = OB – OA
 √ = √

=( ) ( ) 5√ √

Squaring both sides

= ( ) 25( )=

24n2 + 104n + 112 = 0

Using Pythagoras theorem

1digitalteachers.co.ug
 3n2 + 13n + 14 = 0

(3n + 7)(n + 2) = 0

Either n = or n = -2

Hence the values of n are and -2

To show that given points are vertices of a

( ) ( )
right-angle triangle.
Suppose that the points A, B and C are vertices ( ) ( )
of a triangle ABC, to show that ABC is a right
( ) ( ( ))
angled triangle, then by applying the Pythagoras
theorem Since + = = 65, the triangle
PQR is a right angled triangle
Either
or Note: if the triangle is isosceles, then two of
the sides must be equal and for equilateral
or triangle all the sides must be equal

Example 2 Example 3

Prove that the following points are vertices of a Prove that the following points A(1, 2), B(3, 7)
right-angled triangle and C(6, 14) are vertices of an isosceles triangle.

(a) A(2, 3), B(5, 6) and C(8, 3) Solution

Solution

( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
Since , hence, the triangle
Since + = = 36, the triangle ABC, is isosceles triangle.
ABC is a right angled triangle The mid-point of a line segment
(b) P(2, 1), Q(5, -1) and R(9, 5)

Solution

The mid-point , M of a line segment AB with

A(x1, y1) and B(x2, y2) is given as

2digitalteachers.co.ug
 ( ) 5. The points L, M and N have coordinates
(3, 1), ((2, 6) and (x, 5) respectively. Given
Example 4 that the distance LM is equal to the distance
MN, calculate the possible values of x.
(a) Find the coordinates of the midpoint of the
[-3 or 7]
line joining each of the following pairs of
6. Given that the distance between P(r, 4) and
points
Q(2, 3) is equal to the distance between
(i) A(8, 4) and B(2, -4)
R(3, -1) and S(-2, 4). Calculate the possible
Solution
value of r. [-5 or 7]
The midpoint of AB = ( ) 7. A triangle has vertices A(6, 2), B(x, 6) and
= ( ) C(-2, 6). Given that the triangle is isosceles
with AB = BC, Calculate the value of x. [3]
= M(5, 0)
8. F(5, 1), G(x, 7) and H(8, 2) are vertices of a
triangle. Given that the length of the side FG
(ii) P(-6, -2) and Q(-4, -5)
is twice the length of side FH, find the value
Solution
of x. [3 or 7]
The midpoint of PQ = ( ) 9. Given that the distance from A(13, 10) to
= ( ) B(1, y) is three times the distance from B to
= M(-5, -3.5) C(-3, -2), find the value of y. [1 or -8]
10. M(6,5) is the midpoint of a straight line
(b) Find the coordinated of point S given that joining the point A to point B, find the
M(3, -2) is the midpoint of the straight line coordinates of B [10, 7]
joining S to T(9, -2) Gradient of a straight line
Solution
The gradient of a line joining points A(x1, y1) and
The midpoint of PQ = ( )
B(x2, y2) is the measure of steepness of the line
 AB and it is a ratio of the change in y-coordinate
to the change in x-coordinate.

Also i.e., gradient =

y = -2 The gradient is usually denoted by m which may

Hence S(-3, -2) be positive or negative
Exercise 1 Note:
1. Find the distance between each of the (i) If the line slopes downwards from left to
following pairs of points right, the gradient
(a) (-7, 3) and (-2, 5) [√ ]
(b) (2, -3) and (7, 7) [ √ ]
(c) (4, -1) and (-2, 1) [ √ ]
2. Prove that a triangle with vertices (1, 2),
(13, 7) and (6, 14) is isosceles.
3. Find the midpoint of the following point
(a) (2, 1) and (4, 5) [3, 3)
(b) (-1, 4) and ( 3, 1) [1, 3]
(c) (-2, 6) and (0, 2) [-1, 4]
4. Prove that the points A(-2, 0), B(0, 2√ ) and
C(2, 0) are vertices of an equilateral triangle.

3digitalteachers.co.ug
(ii) On the other hand, if the line slopes (i) A(5, 4) and B(6, 8)
upwards from left to right, the gradient is Solution
positive. Angle ( )

( )

(ii) A(-3, -5) and (-4, -2)

Angle ( )
( )
( )
( )

i.e.

Angle of a straight line to the horizontal

Suppose that the line in the second
illustration makes θ is the horizontal as
shown below

Hence the angle which AB makes with

the horizontal is 71.60 with positive x-
axis downwards as shown in the
diagram.

Gradient of parallel lines

If two are parallel, their gradients are equal.

From trigonometry,
=

( )

This means that if m =

In the diagram above L1 is parallel to L2 and both
L1 and L2 have the same gradient.

Example 5 Gradient of perpendicular lines

(a) Find the gradient of the straight line joining If two lines are perpendicular, the product of
each of the following pairs of points. their gradients is -1.
(i) A(7, 4) and B(-1, -2)
Solution
m=
(ii) A(-3, -2) and B(-4, -5)
Solution
( )
m=
( )
(b) Find the angle which the straight line joining If m1 and m2 are the gradients of L1 and L2
each of the following pairs of points makes respectively, then m1 x m2 = -1
with the horizontal

4digitalteachers.co.ug
Example 6 (c) The quadrilateral ABCD has vertices
A(-2, -3), B(1, -1), C(7, -10) and D(2, -9).
(a) Given the points A(2, 3), B(5, 5), C(7,2) and
(i) Prove that Ad is parallel to BC
D(4, 0) ( )
(i) Prove that AB is parallel to DC Gradient of AD, m1 =
( )
( )
Gradient of AB, m1 = Gradient of BC, m2 =
Gradient of DC, m2 = Since the gradient of AD and BC are
Since the gradient of AB and DC are equal, the lines AD and BC are parallel
equal, the lines AB and DC are parallel
(ii) Prove that AC is perpendicular to BD (ii) Prove that AD is perpendicular to BC
( )
Gradient of AC, m1 = Gradient of AD, m1 =
( )
( )
Gradient of BD, m2 = Gradient of BC, m2 =
Since m1 x m2 = -1, AC is perpendicular Since m1 x m2 = -1, AD is perpendicular
BD BC
(b) Prove that the point P(1, 3), Q(3, 4), R(5, 0)
and S(3, -1) form a parallelogram. (iii) Prove that the area of the quadrilateral
ABCD is 32½ sq. units

√( ( )) ( ( ))
√
√( ) ( ( ))
√
Gradient of PQ, m1 = √( ( )) ( ( ))
( )
Gradient of SR, m2 = √
PQ and SR are parallel Since all the sides of a quadrilateral are
different, the figure is a trapezium
√( ) ( ) √

√( ) ( ( )) √
PQ and SR are equal

Gradient of PS, m1 =
) √ (√ √ )
Gradient of QR, m2 =

PS and QR are parallel

So the figure PQRS could either be a (d) A quadrilateral ABCD has vertices A(-2, 1),
rectangle of parallelogram B(0, 4), C(3, 2) and D(1, -1)
For a rectangle (i) Prove that all sides of the quadrilateral
PQ and PS are perpendicular, thus the have the same length.
product of their gradients = -1 Solution
Since the gradient of PQ and x gradient of PS
= √( ( )) ( )
Hence the figure PQRS is a rectangle not a √
parallelogram.

5digitalteachers.co.ug
 √( ) ( ) lines AB, BC and CA. Hence prove that the
√ triangle is right-angled
√( ) ( ) * +
√ 4. The straight line joining the points P(6, 5) to
Q(q, 2) is perpendicular to the straight line
joining point Q to R( 9, -1). Find the value of
√( ) ( ( )) q. [3, 11]
√ 5. Prove that the quadrilateral PQRS with
Hence all the sides of the quadrilateral vertices P(-1, 3), Q(2, 4), R(4, -2) and S(1, -3)
are equal in length is a rectangle and calculate the area [20 sq.
(ii) Prove that AB is parallel to AD units]
6. The four points A(5, 4), B(6,2), C(12, 5) and
) D(11, 7) are vertices of a quadrilateral. Prove
Gradient of AB, m1 =
( )
( )
that the quadrilateral is a rectangle and
Gradient of DC, m2 = calculate its area. [15 sq. units]
Since the gradient of AB and DC are 7. Prove that the points A(2, 3), B(4, 8), C(8, 9)
equal, the lines AB and DC are parallel and D(4, -1)for a trapezium.
8. The quadrilateral ABCD has vertices S(1, 1),
(iii) Prove that AD is parallel to BC T(4, 5), U(12, -1) and V(1, -1) are vertices of
) a quadrilateral STUV.
Gradient of AD, m1 =
( )
(a) Prove that ST is perpendicular to TU, and
Gradient of BC, m2 = that SV is perpendicular to UV.
Since the gradients of AD and BC are (b) Calculate the length of each of the sides
equal, the lines AD and BCC are parallel ST, TU, UV, and VS. [5, 10, 11, 2]
(c) Prove that the area of the quadrilateral
(iv) What is the name of the quadrilateral STUV is 36 square units.
ABCD?
It could be a square or a rhombus 9. The quadrilateral CDEF has vertices, C4, 0),
For square, gradient of AB x gradient of D(8, 4), E(2, -8) and F(0, 2). The points P, Q,
BC = -1 R and S are midpoints of the sides CD, DE, EF
Hence the quadrilateral is a square. and FC respectively. Prove that the
Quadrilateral PQRS is a rhombus and show
that its area is 15 sq. units.
Exercise 2 Equation of a straight line
1. Find the gradient of the straight line of each The general equation of a straight line is given
of the following pairs of points. by y = mx + c, where m = gradient of the line i.e.
(a) (-2, 5) and (5, -3) * + m= and c = y-intercept. For lines passing
(b) (3, 7) and (7, -4) * + through the origin, c = 0; hence y = mx.
(c) (6, 3) and (7, 4) [ ] Example 7
2. Find the angle between a line joining the
following points with the horizontal (a) Find the equation of with gradient = 5 and
(a) (2, 5) and (-3, -2) [54.460] passes through the point
(b) (3, 7) and (-6, 11) [-23.960] (i) A(2, 5)
(c) (5, -3) and (5, 2) [900] Solution
3. A triangle has vertices A(3, -2), B(2, -14) and Method 1
C(-2, -4).Find the gradients of the straight The general equation of a line is y = mx + c

6digitalteachers.co.ug
 Substituting for m = 5 and points of A
5 = 5(2) + c
c = -5
Hence the line is y = 5x – 5
Method 2
Let B(x, y) lie on the line
From m =
5= Gradient of AB, m1 =
y – 5 = 5(x – 2)
y = 5x – 5 Gradient of MP, m2 =

But m1 x m2 = -1
(ii) P(2, 7)
Using y = mx + c ( )
Substituting for m = 5 and points of P
7 = 5(2) + c 4(y – 2) = 3(x – )
c = 17
Hence the line is y = 5x +17 8y -16 = 6x – 3

8y – 6x = 13
(b) Find the equation of a straight line joining
the following points (ii) P(-3, -5) and Q(3, -7)
(i) A(0, 5) and B(3, 4)
Solution Solution
Gradient = Let M be the midpoint of AB
Substituting coordinates for A in general  M( =M( )
equation
Let R(x, y) lie on the perpendicular bisector
5= ( ) => C = 5
Hence equation of the line is y =
(ii) P(-2, -5) and Q(3, -7)
Solution
( )
Gradient =
( )
Substituting coordinates for P in general
equation
-5 = ( ) => C = -5 - = - Gradient of PQ, m1 =
( )

Hence equation of the line is y =

( )
(c) Find the equation of a perpendicular Gradient of MR, m2 =
bisector of the line joining the following
But m1 x m2 = -1
points
(i) (2, 0) and (-1, 4) ( )
Solution
Let M be the midpoint of AB 5(y+6) =2x
 M( =M(0.5, 2) 5y – 2x + 30 = 0

Let P(x, y) lie on the perpendicular bisector (d) A straight line, L passes through point (-2, 1)
and makes an angle of 450 with the
horizontal.

7digitalteachers.co.ug
(i) Find the equation of the line L. ( ) ( )
d =| | =4√ units
√ √
Solution
(b) (7, -4) from 3x – 5y = 7
Solution
Rearranging the equation
3x – 5y – 7 = 0
a =3, b = -5 and c = -7
Substituting for (x, y) = (7, -4)
( ) ( )
d =| | =√ units
√ ( ) √

(c) (8, -5) from y = 3x -1

Solution
Gradient of the line L = tan 450 = 1
Rearranging the equation
 m=2
3x – y – 1 = 0
Substituting the coordinates of the point in a =3, b = -1 and c = -1
the general equation, y = mx + c Substituting for (x, y) = (8, -5)
( ) ( )
1 = 1(-2) + c => that c = 3 d =| | = units
√ ( ) √

Hence the equation of the line L is y = x + 3

Intersection of two lines
(ii) Given that L intersects the x-axis at A and The point of intersection of two or more
the y-axis at B find the distance AB lines is obtained by solving the equations of
Solution the lines simultaneously.
At A, y = 0; x = -3
Hence coordinates of A(-3, 0)
At B x = 0 y = 3
Hence coordinates of B(0, 3)
√( ( )) ( )
√ units

Perpendicular distance of a point to a line

Just like in vectors, the shortest distance of a
point from a given line is the perpendicular
distance of a point from the line Example 9
Suppose that the equation of the line is in
(a) Find the coordinates of the point of
the form ax + by + c = 0
intersection of each of the following pairs of
The perpendicular distance of a point
straight lines
P(x1, y1)=d =| | (i) y = x + 3 and y = 4x + 6
√
Solution
Example 8
Let y = x + 3 ……………………(1)
Find the perpendicular distance of the following And y = 4x + 6 …………………(2)
points from the given lines Equating (1) and (2)
x + 3 = 4x + 6; x = -1
(a) (-2, 6) from x + y + 4 = 0
substituting x I (1)
Solution
y = 4(-1) + 6; y = 2
a =1, b = 1 and c = 4
Hence the point of intersection is (-1, 2)
Substituting for (x, y) = (-2, 6)
(ii) 2x - 3y = 7 and 3x – 7y = 13

8digitalteachers.co.ug
 Let 2x – 3y = 7 ……………………(1) 2y + x = 8
And 3x – 7y = 13 …………………(2) At the point, P of intersection, the two
Equating (1) and (2) lines are equal
x + 3 = 4x + 6; x = -1 2x – 1 = y ………………………..(1)
3eqn,(1) – 2eqn. (2) 2y + x = 8 ……………………….. (2)
5y = -5 => y = -1 Substituting for y in eqn. (1) into
Substituting for y into eqn. (1) equation (2)
2x – 3(-1) = 7; x = 2 2(2x-1) + x = 8
Hence the point of intersection (2, -1) x=2
Substituting for x in eqn. (1)
(b) (i) Find the equation of the straight line L, y = 2(2) – 1 = 3
which passes through the point P(2, 4) and Hence point of intersection P (2, 3)
perpendicular to the line 5y + x = 7 (ii) Find an equation of N and hence find the
Solution coordinates of point Q where the Line L
Let Q(x, Y) lie on the same line and Line N intersect
Gradient of PQ, m1 = Let D (x, y) lie on N

From5y + x = 7; y = Gradient BD, m1 =

From line 2y + x = 8
Gradient, m2 =
y=
But for perpendicular lines, m1 x m2 = -1
Gradient m2 =
 ( )
For parallel lines m1 = m2.
4 = 5(x – 2)
6 
(ii) Given that the line L meets line y = x + 6
at point S. find the coordinates of S. 2y + 3x = 3
Solution At point Q, Line L = line N
At S the two equations are equal 2x – 1 = y …………………..(1)
 6 = x + 6; x = 3 2y + x = 3 …………………….(2)
Substituting x in y = x + 6 Substituting for y in eqn. (2)
y=3+6=9 2(2x -1) + x = 3
Hence coordinates of S are (3, 9) 5x – 2 = 3
(c) The line L has equation 2x – y – 1 = 0. The x=1
line M passes through point A(0, 4) and is Substituting for x in eqn. (1)
perpendicular to the line L. The line N passes y = 2(1) -1 = 1
through point B(3, 0) and is parallel to M. hence Q(1, 1)
(i) Find an equation of M and show that he (iii) Prove that
line L and M intersect at the point
A(0, 4). √( ) ( ) √
Solution √( ) ( ) √
Let K(x, y) lie on line M
√( ) ( ) √
Gradient of AK, m1 =
Hence = =
From 2x – y = 1; y = 2x -1
Gradient, m2 = 2 Angle between two lines
For perpendicular line m1 x m2 = -1 Suppose that two lines y = m1x + c1 and
 ( ) y = m2x + c2 are inclined at angle α and β
2(y – 4) = -x respectively as shown below

9digitalteachers.co.ug
 From θ = ( )
( )
θ= ( ) ( )
( )

= ( ) =46.80
(iii) y = 4 and 3y + 2x – 6 = 0
Solution
For y = 4, => m1 =
The sum of two interior angles = opposite For 3y + 2x – 6 = 0
exterior angle
y= => m2 =
 θ+β=α
θ=α–β From θ = ( )
tan θ = tan (α – β) ( )
θ= ( ) ( )
= ( )

= = ( ) =33.70

θ= ( )
(b) Calculate the area of the triangle which has
sides given by the equations 2y – x = 1,
Hence the acute angle between two y + 2x = 8 and 4y + 3x = 7
intersecting lines is ( ) Solution
Note that the angle between parallel lines is
zero while that between perpendicular line
is 900.

Example 10

(a) Find the acute angle between the following

pairs of lines
(i) 4y – 3x = 6 and 2y + x = 3
Solution At point P
For 4y – 3x = 6 4y + 3x = 7 …………………………………(1)
y= => m1 = 2y – x = 1 …………………………………...(2)
Eqn. (1) – 2eqn.(2)
For 2y + x = 3
10y = 10; y = 1
y= => m2 =
Substituting for y in eqn. (2)
From θ = ( ) x = 2(1) – 1 = 1
( ) Hence P(1, 1)
θ= ( ) ( )
( )
At point Q
= =63.40
y + 2x = 8 …………………………………(1)
2y – x = 1 …………………………………...(2)
(ii) 2x – 5y = 15 and 3y + 2x = 6
Eqn. (1) + 2eqn.(2)
Solution
5y = 10; y = 2
For 2x – 5y = 15
Substituting for y in eqn. (2)
y= => m1 =
x = 2(2) – 1 = 3
For 3y + 2x = 6 Hence Q(3, 2)
y= => m2 =

10digitalteachers.co.ug
 At point R √ √ √ √
Area = √( )( )( )( )
y + 2x = 8 …………………………………(1)
4y + 3x = 7 ……………………………....(2) = 5sq. units
3Eqn. (1) - 2eqn.(2) Exercise 3
-5y = 10; y = -2 1. Find the gradient of each of the following
Substituting for y in eqn. (1) straight lines
2x = 8 + 2 =10 => x = 5 (a) y = 4x -2 [4]
Hence r(-2, 5) (b) y = 2x + 3 [3]
Finding dimensions (c) y = 2 – 5x [-5]
√( ) ( ) (d) * +
√( ) ( ) √ 2. Find the equation of the straight line that
√( ) ( ) √ has the following properties
Finding <QPR (a) Gradient 1 and passes through (2, 4)
For 4y + 3x = 7 [y = x + 2]
y= => m1 = (b) gradient and passes through (2, 5)
For 2y – x = 1 * +
y= => m2 = 3. Find the equation of a straight line that has
From θ = ( ) the following properties
(a) Passes through (-2, 3) and parallel to
( )
<QPR = ( ) ( ) y = 5x + 4
( )
[ ]
= ( ) (b) Passes through (6, -2 and is
perpendicular to y = -3x + 4
* +
(c) Passes through ( ) and is
perpendicular to 3y + 10x – 8 = 0
[ ]
4. Find the equation of a straight line joining
the following pairs of points
(a) (2, 4) and (-1, 0) [3y = 4x +4]
(b) (-4, 1) and(6, 2) [10y = x + 14]
(c) (3, 4) and (-1, 4) [y =4]
5. Find the equation of the perpendicular
bisector of the straight lines joining each of
Method 1 using sine rule the following pairs of point
Area of PQR = (a) (5, 5) and (2, -2) [4y+ 2x = 11]
= √ sq. units (b) (-1, 4)and (3, 3) [2y – 8x + 1 = 0]
√
(c) (3, 2) and (-4, 1) [y = -7x – 2]
Method 2 using Heron’s formula
6. Find the equation of a straight line:
(a) L1 which is perpendicular bisector of
Area of PQR = √ ( )( )( )
points A(-2, 3) and B(1, -5)
Where a, b, and c are sides of a triangle and [16y – 6x + 13 = 0]
s= ( ) (b) L2 which is a perpendicular bisector of
s= ( √ √
√ the points B(1, -5) and (17, 1)
[3y + 8x – 60 = 0]
(c) Show that L1 is perpendicular to L2.

11digitalteachers.co.ug
7. The perpendicular bisector of a straight line [y = 5x – 6]
joining the points (3, 2) and (5, 6) meet the (b) Given that the line L meets the line
x-axis at A and they-axis at B. Prove that the y = x + 6 at point S, find the coordinates
distance AB is equal to √ . of point S.[3, 9]
8. A is a point (1, 2) and B is a point (7, 4). The 13. Calculate the area of the triangle which has
straight line L1 passes through B and is sides given by the equations 2y – x = 1,
perpendicular to AB; the straight line L2 y + 2x = 8 and 4y + 3x = 7 [5sq. units]
passes through A and is also perpendicular 14. The point A has coordinates (2, -5). The
to AB. The line L1 meets the x-axis at P and straight line 3x + 4y – 36 = 0 cuts the x-axis
the y-axis at Q. Line L2 meets the x-axis at R at B and the y – axis at C. Find
and the y-axis at S. (a) The equation of the line through A
(a) Find the equations of each of L1 and L2. which is perpendicular to the line BC.
[y = -3x + 25, y = -3x + 5] [4x – 3y = 23]
(b) Calculate the area of the triangle OPQ. (b) The perpendicular distance from the line
* + BC. [10]
(c) The area of the triangle ABC. [75 sq.
(c) Calculate the area of the triangle ORS.
units]
* +
(d) Find the area of the trapezium PQSR. Locus
[100 sq. units] A locus is the set of all points in a plane that
9. P is the point with coordinates (2, 1) and L is satisfy some condition. For example the
the straight line which is perpendicular to locus of points equidistant from two given
OP and which passes through P. points say A and B is a perpendicular
(a) Find the equation of L. [y = -2x + 5] bisector of AB.
(b) Given that line L meets the x-axis at A
and y-axis at B. calculate Locus can be expressed in terms of the
(i) the area of the triangle OAP. [1.25] Cartesian coordinates (x, y) of the form (r, θ)
(ii) The area of the triangle OBP [5] Example 11
(iii) Find the ratio of the area OAP to
that of OBP [1:4] (a) Find the locus of point P(x, y) that is
10. Find the shortest distance between each of equidistant from the point A(2, -1) and
the following B(-1, 2)
(a) The point (2, 4) and the line Solution
3x – 4y + 8 = 0 * + Method 1

(b) The point (5, -1) and the line

12x + 5y – 3 = 0 [4]
(c) The point (9, -3) and the line y = x. [6√ ]
11. Find the coordinates of the point of
intersection of each of the following pairs of
straight lines
(a) y = 2x + 3 and y = 4x + 1 [1, 5]
(b) y = x + 3 and y = 4x + 6 [-1, 2] i.e.
(c) 2x – 3y = 7 and 3x – 7y = 13 [2, -1] ( ) ( ) ( ) ( )
(d) x + 3y – 2 = 0 and 3x + 5y – 8 = 0 * + y=x
12. Find the equation of the straight line L,
which passes through the point (2, 4) and
perpendicular to the line 5y + x = 7

12digitalteachers.co.ug
Method 2 The centre of the circle is (1, -2) and the
radius= √ = 2√ units
P(x, y) (iii) Where does P cut the line x = 3?
Solution
Substituting x = 3 in the equation
(x – 1)2 + (y + 2)2 = 20
A(-1, 2) B(2, -1) ((3 – 1)2 + (y + 2)2 = 20
M( )
(y + 2)2 = 16 i.e. y + 2 =
y = -2 i.e. y = -6 or y = 2
Gradient of AB = P cuts the line x = 3 at the point (3, 2)
Since AB and MP are perpendicular, the product and (3, -3)
of their gradients = -1
(d) Find the locus of P(x, y) if its distance from
Gradient of MP= A(-1, 1) is equal to the distance form the line
y=x 2x – y = 1
Solution
(b) Find the locus of a point which moves so
that of the squares of its distance from
points A(-2, 0) and B(2, 0) is 25 units.

| |
√( ) ( ) =
√( ( ) )
= 25  [( ) ( ) ] ( )
( ) ( ) = 25 5x +10x + 5 + 5y – 10y + 5 = 4x(y+1)+ (y+1)2
2 2

x2 + 4y2 + 4xy + 14x – 12y + 9 = 0 is the locus

(e) A point R moves so that its distance from
(c) The locus of P(x, y) is such that the distance point (2, 0) is twice its distance from (0, -1).
OP is half the distance PR, where O is the Show that the locus of R is a circle and
origin and R is the point (-3, 6) determine its radius and its centre.
(i) Show that the locus of P describes a Solution
circle in the x-y plane Let P (2, 0), Q(0, 1) and R (x, y)
Solution Given PR = 2QR => PR2 = 4QR2
Given OP = i.e. 4OP2 = PR2 (x – 2)2 + y2 = 4[x2 + (y+ 1)2]
4(x2+ y2) = (x+ 3)2 + (y – 6)2 x2 - 4x + 4 +y2 = 4x2 + 4y2 + 8y + 4
4x2 + 4y2 = x2 + 6x + 9 + y2 – 12y + 36 3x2 + 3y2 + 4x + 8y = 0 hence a circle
x2 + y2 – 2x + 4y – 15 = 0 (a circle in x – y  + =0
plane)
( ) ( )
(ii) Determine the radius of the circle and
the centre of the circle. The centre is at ( ) and the radius
Solution √
x2 + y2 – 2x + 4y = 15 =√
After completing squares we have
(x – 1)2 + (y + 2)2 = 20

13digitalteachers.co.ug
(f) Find the locus of a point P(x, y) whose
distance from the point A(3, -2) is always
(x -1)(2 +y) =y(x – 1)
5units.
y – x = -1 (which is a line)
Solution
(i) A point P moves such that its distance from
AP = 5 => AP2 = 25
two points A(-2, 0) and B (8,6) are in ratio
i.e. (x – 3)2 + (y + 5)2 = 25
AP: PB = 3:2. Show that the locus of P s a
x2 + y2 – 6x + 4y – 12 = 0
circle. (05marks)
The locus is a circle with centre (3, -2) and ̅̅̅̅
radius 5 units. ̅̅̅̅
=> ̅̅̅̅ ̅̅̅̅
(g) A is a point on the x-axis and C is a point √( )
(2, 3). The perpendicular to AC through C √( ) ( )
meets the y-axis at B. Find the locus of the Squaring both sides
midpoint of AB. 4( )
Solution = 9(x2 -16x + 64 + y2 -12y + 36)
Let A(a, 0), B(0, b) and M(x, y) is the 4x2 + 16x + 16 + 4y2
midpoint of AB = 9x2 – 144x + 900 + 9y2 – 108y
 M( ) i.e. M( ) 5x2 + 5y2- 160x -108y + 884 = 0
Since AC is perpendicular BC
 Exercise 4
1. Find the equation to the locus of a point
( 3)=2(2-a) …………………….(i)
which moves so that its distance from the
point (3, 4) is always 5 units.
[x2 + y2 – 6x – 8y = 0]
2. A and B are points (1, 0) and (7, 8)
respectively. A point P moves so that the
angle APB is a right angle. Find the equation
of locus of P.

[x2 + y2 – xx – 8y + 7 = 0]

3. A variable point P is given by parametric

Now at M , x = and y = equation x = ct, y = . Show that the locus of
 a = 2x and b = 2y P is xy = c2.
Substituting into eqn. (i) 4. Find the locus of point P which is equidistant
3(2y - 3) =2(2 – 2x) from the line x = 2and the circle x2 + y2 = 1.
i.e. 4x + 6y= 13 [x2 + 6x – 9 = 0]
5. A Point P is twice as far from the line x + y =
(h) P is a variable point given by parametric 5 and from the point (3, 0). Find the locus of
equations P.
and
[7x2 + 7y2 – 38x + 10y – 2xy + 47 = 0]
Show that the locus of P is a straight line
Solution 6. Find the Cartesian equation of a curve given
By eliminating t parametrically by and
From ;t= [y – x + 1= 0]
7. P(3, 2) and Q(5, 6) are two fixed points and R
From ;t= . moves so that angle PRQ is right angle. Show
that the locus of R is given by the equation
Equating t
x2 + y2 – 9x – 6y + 26 = 0

14digitalteachers.co.ug
8. Show that the locus of P is , radius of the circle
√
given that PA + PB = 4 where A and B are the * ( ) +
points (1, 0) and (-1, 0) respectively 9. Find the locus of the point P(x, y) which
moves such that its distance from the point
Exercise 5 (Topical question) S(-3, 0) is equal to its distance from a fixed
line x = 3 [y2 + 12x = 0]
1. ABCD is a quadrilateral with A(2, -2), B(5, -1), 10. Given the vector a = i – 3j + 3k and
C(6, 2) and D(3, 1). Show that the b = -i – 3j + 2k. find
quadrilateral is a rhombus. (j) acute angle between vectors a and b
2. PQRS is a quadrilateral with vertices P(2, -1), [30.860]
Q(4, -1) and S(2, 1). Show that the (ii) equation of the plane containing a and b
quadrilateral is a rhombus [-3x+ 5y +6z = 0]
3. The Locus of P is such that the distance OP is 11. The points A and B lie on the positive sides
half the distance PR, where O is the origin of the x-axis and y-axis respectively. If the
and R id the point (-3, 6). length of AB is 5 units and angle OAB is θ,
(a) Show that the locus of P describes a where O is the origin, find the equation of
circle in x – y plane the line AB (leave θ in your answer)
[x2 + y2 – 2x + 4y – 15 = 0 (a circle in x – y [y = -xtanθ + 5sinθ]
plane)] 12. (a) Find the equation of the locus of a point
(b) Determine the centre and radius of the which moves such that its distance from
circle. D(4, 5) is twice its distance from the origin.
[The centre of the circle is (1, -2) and the [8x2 + 8y2 + 8x + 10y – 41 = 0]
radius= √ = 2√ units] (b) the line y = mx intersects the curve 2x2 –
(c) Where does P cuts the line x = 3 x at point A and B. Find the equation of the
[(3, 2) and (3, -3)] locus of point P which divides AB in the ratio
4. A Point P is twice as far from the line 2:5. [y = 7x2 –x]
x + y = 5 as from the point (3, 0). Find the
locus of P.
[7x2 + 7y2 -38x + 10y + 47 = 0]
5. Find the locus of point P which is equidistant
from the line x = 2and the circle x2 + y2 = 1.
[x2 + 6x – 9 = 0]
6. The points R(2, 0) and P (3. 0)lie on the x-
axis and Q(0, -y) lie on the y-axis. The
perpendicular from the origin to RQ meets
PQ at S(X, -Y). Determine the locus Of S in
terms of X and Y. [2X2+ 3Y2 – 6X = 0]
7. The point A(2, 1), P(α, β) and point B(1, 2) lie
in the same plane. PA meets the x-axis at
point (h, 0) and PB meets the y-axis at point
(0, k). Find h and k in terms of α and β.
* +
8. A is a point (1, 3) and B is a point (4, 6). P is a
variable point which moves in such a way
Thank you
that . Show that the locus
of P describes a circle. Find the centre and Dr. Bbosa Science

15digitalteachers.co.ug