Coordinate Geometry
by
S. L. Loney
Geometry is concerned with properties of space that are related with
distance, shape, size, and relative position of figures.
Algebra is the study of mathematical symbols and the rules for
manipulating these symbols appearing in equations describing relationships
between variables.
Rene Descartes coupled the branch of geometry with that of algebra
and thereby the study of geometry could be simplified to a very large extent. Of
course, this hindered the development of abstract geometry, just like the
availability of digital computers blocked the growth of mathematics.
22. To find the coordinates of the point this divides in a given ratio 𝑚1 ∶ 𝑚2 the
line joining two given points 𝑥1,𝑦1 𝑎𝑛𝑑 𝑥2,𝑦2 .
Questions (10)
Find the distance between the following pairs of points.
2, 3 𝑎𝑛𝑑 5, 7
1, −7 𝑎𝑛𝑑 −1, 5
−3, −2 and −6, 7
𝑎, 0 𝑎𝑛𝑑 0, 𝑏
𝑏 + 𝑐, 𝑐 + 𝑎 𝑎𝑛𝑑 𝑐 + 𝑎, 𝑎 + 𝑏
𝑎 cos 𝛼 , 𝑎 sin 𝛼 𝑎𝑛𝑑 𝑎 cos 𝛽 , 𝑎 sin 𝛽
𝑎 𝑚12 , 2𝑎𝑚1 𝑎𝑛𝑑 𝑎 𝑚22 , 2𝑎𝑚2
Lay down in a figure the positions of the points 1, −3 and −2, 1 and
prove that the distance between them is 5.
9. Find the value of 𝑥1 if the distance between the points 𝑥1 , 2 𝑎𝑛𝑑 3, 4
be 8.
10.A line is of length 10 and one end is at the point 2, −3 ; if the abscissa of
the other end be 10, prove that its ordinate must be 3 or -9.
1.
2.
3.
4.
5.
6.
7.
8.
Questions (8)
1. Prove that the points 2𝑎, 4𝑎 , 2𝑎, 6𝑎 , and 2𝑎 + 3𝑎, 5𝑎 are the
vertices of an equilateral triangle whose side is 2a.
2. Prove that the points −2, −1 , 1, 0 , 4, 3 , 𝑎𝑛𝑑 1, 2 are at the
vertices of a parallelogram.
3. Prove that the points 2, −2 , 8, 4 , 5, 7 , 𝑎𝑛𝑑 −1, 1 are at the
angular points of a rectangle.
4. Prove that the points −
5.
6.
7.
8.
1
14
,
39
14
is the centre of the circle
circumscribing the triangle whose angular points are 1, 1 , 2, 3 ,
𝑎𝑛𝑑 −2, 2 .
Find the coordinates of the point which divides the line joining the points
1, 3 𝑎𝑛𝑑 2, 7 in the ratio 3: 4.
Find the coordinates of the point which divides the same line in the ration
3: − 4.
Find the coordinates of the point which divides, internally and externally,
the line joining −1, 2 to 4, 5 in the ratio 2: 3.
Find the coordinates of the point which divides, internally and externally,
the line joining −3, −4 𝑡𝑜 −8, 7 in the ratio 7: 5.
Questions (3)
1. The line joining the points 1, −2 and −3, 4 is trisected; find the
coordinates of the points of trisection.
2. The line joining the points −6, 8 𝑎𝑛𝑑 8, −6 is divided into four equal
parts; find the coordinates of the points of section.
3. Find the coordinates of the points which divide, internally and externally,
the line joining the point 𝑎 + 𝑏, 𝑎 − 𝑏 to the point 𝑎 − 𝑏, 𝑎 + 𝑏 in the
ratio 𝑎: 𝑏.
Questions (9)
Find the area of the triangles the coordinates of whose angular points are
respectively
1.
2.
3.
4.
5.
6.
7.
8.
1, 3 , −7, 6 𝑎𝑛𝑑 5, −1
0, 4 , 3, 6 𝑎𝑛𝑑 −8, −2
5, 2 , −9, −3 𝑎𝑛𝑑 −3, −5
𝑎, 𝑏 + 𝑐 , 𝑎, 𝑏 − 𝑐 𝑎𝑛𝑑 – 𝑎, 𝑐
𝑎, 𝑐 + 𝑎 , 𝑎, 𝑐 𝑎𝑛𝑑 −𝑎, 𝑐 − 𝑎
𝑎 cos 𝜙1 , 𝑏 sin 𝜙1 , 𝑎 cos 𝜙2 , 𝑏 sin 𝜙2 𝑎𝑛𝑑 𝑎 cos 𝜙3 , 𝑏 sin 𝜙3
𝑎𝑚1 2 , 2𝑎𝑚1 , 𝑎𝑚2 2 , 2𝑎𝑚2 𝑎𝑛𝑑 𝑎𝑚3 2 , 2𝑎𝑚3
𝑎𝑚1 𝑚2 , 𝑎 𝑚1 + 𝑚2 , 𝑎𝑚2 𝑚3 , 𝑎 𝑚2 + 𝑚3 and 𝑎𝑚3 𝑚1 ,
𝑎 𝑚3 + 𝑚1
9. 𝑎𝑚1 ,
𝑎
𝑚1
, 𝑎𝑚2 ,
𝑎
𝑚2
, 𝑎𝑛𝑑
𝑎𝑚3 ,
𝑎
𝑚3
Questions (3)
Prove (by showing that the area of the triangle formed by them is zero) that
the following sets of three points are in a Straight line:
1. 1, 4 , 3, −2 , 𝑎𝑛𝑑 −3, 16
1
2. − , 3 , −5, 6 , 𝑎𝑛𝑑 −8, 8
2
3. 𝑎, 𝑏 + 𝑐 , 𝑏, 𝑐 + 𝑎 , 𝑎𝑛𝑑 𝑐, 𝑎 + 𝑏
Locus. Equation to a Locus
• Article 36 When a point moves so as always to
satisfy a given condition or conditions, the path it
traces out is called its Locus under these
conditions.
For example, suppose O to be given point in
the plane of the paper and that a point P is to
move on the paper so that its distance from O
shall be constant and equal to a. It is clear that all
the positions of the moving point must lie on the
circumference of a circle whose centre is O and
radius is a. The circumference of this circle is
therefore the “Locus” of P when it moves subject
to the condition that its distance from O shall be
equal to the constant distance a.
Equation to a curve
• Definition
The equation to a curve is the relation
which exists between the coordinates of any
point on the curve, and which holds for no
other points except those lying on the curve.
Straight Line
Article 46 To find the equation to a straight line
which is parallel to one of the coordinate axes.
Let CL be any line parallel to the axis of y and
passing through a point C on the axis of x such
that OC=c.
Let P be any point this line whose coordinates are x
and y. Then the abscissa of the point P is always c,
so that
x=c
……..(1)
Straight line parallel to the axes
Y
L
P
O
C
Continued
The expression (1) is true for every point on the
line CL (produced indefinitely both ways), and for
no other point , is, by article 42, the equation to
the line.
It should be noted that the equation does not
contain the coordinate y.
Similarly the equation to straight line parallel to the
axis of x is y=d.
Corollary: The equation to the axis of x is y=0.
The equation to the axis of y is x=0.
47. To find the equation to a straight line which cuts off a
given intercept on the axis y and is inclined at a given angle to
the axis of x.
Y
L
P
C
N
O
M
L’
X
Let the given intercept be c and let the given angle be 𝛼.
Let C be a point on the axis of y such that OC is c. Through C draw a straight line
LCL’ inclined at an angle 𝛼 = tan−1 𝑚 to the axis of x, so that tan 𝛼 = 𝑚.
The straight line LCL’ is therefore the straight line required, and we have to find
the relation between the coordinates of any point P lying on it.
Draw PM perpendicular to OX to meet in N a line through C parallel to OX.
Let the coordinates of P be x and y, so that 𝑂𝑀 = 𝑥 𝑎𝑛𝑑 𝑀𝑃 = 𝑦.
Then 𝑀𝑃 = 𝑁𝑃 + 𝑀𝑁 = 𝐶𝑁 tan 𝛼 + 𝑂𝐶 = 𝑚𝑥 + 𝑐,
i.e.
𝑦 = 𝑚𝑥 + 𝑐
This relation being true for any point on the given straight line is, by Art 42, the
equation to the straight line.
Corollary: The equation to nay straight line passing through the origin, i.e. which
cuts off a zero intercept from the axis of y, is found by putting 𝑐 = 0 and hence is
𝑦 = 𝑚𝑥.
Example
The equation to the straight line cutting off an intercept 3 from the negative
direction of the axis of y, and inclined at 120° to the axis of x, is
𝑦 = 𝑥 tan120°+ −3
i.e.
𝑦 = −𝑥 3 − 3
i.e.
𝑦 + 𝑥 3 + 3 = 0.
50. To find the equation to the straight line which cuts off
given intercepts a and b from the axes.
Y
B
P
O
M
A
X
Let A and B be on OX and OY respectively, and be such that 𝑂𝐴 = 𝑎 𝑎𝑛𝑑 𝑂𝐵 =
𝑏. Join AB and produce it indefinitely both ways. Let P be any point 𝑥, 𝑦 on this
straight line, and draw PM perpendicular to OX.
We require the relation that always hold between 𝑥 and 𝑦, so long as P lies on AB.
By geometry, we have 𝑠𝑖𝑛𝑐𝑒 ∆𝐵𝑂𝐴 𝑎𝑛𝑑 ∆𝑃𝑀𝐴 𝑎𝑟𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠
𝑂𝑀 𝑃𝐵
𝑀𝑃 𝐴𝑃
= , 𝑎𝑛𝑑
=
𝑂𝐴 𝐴𝐵
𝑂𝐵 𝐴𝐵
i.e.
𝑂𝑀 𝑀𝑃 𝑃𝐵 + 𝐴𝑃
+
=
=1
𝑂𝐴 𝑂𝐵
𝐴𝐵
𝑥 𝑦
𝑎
+ =1
𝑏
This is therefore the required equation; for it is the relation that holds
between the coordinates of any point lying on the given straight line.
51. Example: Find the equation to the straight line passing through the point
3, −4 and cutting off intercepts, equal but of opposite signs, from the axes.
The equation to the straight line is then
𝑥
𝑦
+
=1
𝑎 −𝑎
𝑥−𝑦 =𝑎
i.e.
… (1)
Since, in addition, the straight is to go through the point 3, −4 , these
coordinates must satisfy (1), so that
3 − −4 = 𝑎
and therefore
𝑎 = 7.
The required equation is therefore
𝑥 − 𝑦 = 7.
62. To find the equation to the straight line which
passes through the two given points (x’, y’) and (x’’, y’’).
Y
(x’’, y’’)
(x’, y’)
O
X
By Art. 47, the equation to any straight line is
𝑦 = 𝑚𝑥 + 𝑐
….(1)
By properly determining the quantities 𝑚 𝑎𝑛𝑑 𝑐 we can make (1)
represent nay straight line we need.
If (1) pass through the point 𝑥 ′ , 𝑦′ , we have
𝑦 ′ = 𝑚𝑥 ′ + 𝑐
… … (2)
Substituting for 𝑐, from (2), the equation (1) becomes
𝑦 − 𝑦 ′ = 𝑚 𝑥 − 𝑥′
… … (3)
This is the equation to the line going through 𝑥 ′ , 𝑦′ making an angle
tan−1 𝑚 with OX. If in addition (3) passes through the point 𝑥 ′′ , 𝑦′′ ,
then
𝑦 ′′ − 𝑦 ′ = 𝑚 𝑥 ′′ − 𝑥 ′
This yields
𝑚=
𝑦 ′ ′ −𝑦 ′
𝑥 ′′ −𝑥 ′
.
Substituting this value in (3), we get as the required equation
′′
𝑦
− 𝑦′
′
𝑦 − 𝑦 = ′′
𝑥 − 𝑥′
𝑥 − 𝑥′
Questions (6)
Find the equation to the straight line
1. Cutting off an intercept unity from the positive direction of the
axis of 𝑦 and inclined at 45° to the axis of 𝑥.
2. Cutting off an intercept −5 from the axis of 𝑦 and being equally
inclined to the axes.
3. Cutting off an intercept 2 from the negative direction of the axis
of 𝑦 and inclined at 30° to OX.
4. Cutting of an intercept −3 from the axis of 𝑦 and inclined at an
3
angle tan−1 to the axis of 𝑥.
5
5. Cutting of intercepts 3 𝑎𝑛𝑑 2 from the axes.
6. Cutting of intercepts −5 𝑎𝑛𝑑 6 from the axes.
Questions (4)
1. Find the equation to the straight line which passes through the
point 5, 6 and intercepts on the axes
[1].
Equal in magnitude and both positive,
[2].
Equal in magnitude but opposite in sign.
2. Find the equation to the straight lines which pass through the
point 1, −2 and cut off equal distances from the two axes.
3. Find the equation to the straight line which passes through the
given point 𝑥 ′ , 𝑦′ and is such that the given point bisects the
part intercepted between the axes.
4. Find the equation to the straight line which passes through the
point −4, 3 and is such that the portion of it between the axes
is divided by the point in the ratio 5 ∶ 3.
Questions (7)
Find the equations to the straight lines passing through the following
pairs of points.
1.
2.
3.
4.
5.
6.
0, 0 𝑎𝑛𝑑 2, −2
3, 4 𝑎𝑛𝑑 5, 6
−1, 3 𝑎𝑛𝑑 6, −7
0, 𝑎 𝑎𝑛𝑑 𝑏, 0
𝑎, 𝑏 𝑎𝑛𝑑 𝑎 + 𝑏, 𝑎 − 𝑏
𝑎𝑡1 2 , 2𝑎𝑡1 𝑎𝑛𝑑 𝑎𝑡2 2 , 2𝑎𝑡2
7. 𝑎𝑡1 ,
𝑎
𝑡1
𝑎𝑛𝑑 𝑎𝑡2 ,
𝑎
𝑡2
Questions (6)
Find the equations to the straight lines passing through the following
pairs of points.
1. 𝑎 cos 𝜙1 , 𝑎 sin 𝜙1 𝑎𝑛𝑑 𝑎 cos 𝜙2 , 𝑎 sin 𝜙2
2. 𝑎 cos 𝜙1 , 𝑏 sin 𝜙1 𝑎𝑛𝑑 𝑎 cos 𝜙2 , 𝑏 sin 𝜙2
3. 𝑎 sec 𝜙1 , 𝑏 tan 𝜙1 𝑎𝑛𝑑 𝑎 sec 𝜙2 , 𝑏 tan 𝜙2
Find the equations to the sides of the triangles the coordinates of
whose angular points are respectively.
4. 1, 4 , 2, −3 , 𝑎𝑛𝑑 −1, −2
5. 0, 1 , 2, 0 , 𝑎𝑛𝑑 −1, −2
6. Find the equations to the diagonals of the rectangle the equations
of whose sides are 𝑥 = 𝑎, 𝑥 = 𝑎′ , 𝑦 = 𝑏, 𝑎𝑛𝑑 𝑦 = 𝑏′.
66. To find the angle between two given straight lines
Y
A
C2
C1
L2
L1
O
X
Let the two straight lines be AL1 and AL2, meeting the axes of 𝑥 𝑖𝑛 𝐿1
and L2. Let their equations be
𝑦 = 𝑚1 𝑥 + 𝑐1 𝑎𝑛𝑑 𝑦 = 𝑚2 𝑥 + 𝑐2
… … … . (1)
We know that
tan 𝐴𝐿1 𝑋 = 𝑚1 , 𝑎𝑛𝑑 tan 𝐴𝐿2 𝑋 = 𝑚2
∠𝐿1 𝐴𝐿2 = ∠𝐴𝐿1 𝑋 − ∠𝐴𝐿2 𝑋.
Now
∴
tan ∠𝐿1 𝐴𝐿2 = tan ∠𝐴𝐿1 𝑋 − ∠𝐴𝐿2 𝑋
tan 𝐴𝐿1 𝑋−tan 𝐴𝐿2 𝑋
1+tan 𝐴𝐿1 𝑋∙tan 𝐴𝐿2 𝑋
Hence the required angle ∠𝐿1 𝐴𝐿2 =
=
𝑚 1 −𝑚 2
1+𝑚 1 𝑚 2
−1 𝑚 1 −𝑚 2
tan
1+𝑚 1 𝑚 2
… . (2)
67. To find the condition that two straight lines may be parallel.
Two straight lines are parallel when the angle between them is zero
and therefore the tangent of this angle is zero. This gives
𝑚1 − 𝑚2
tan 0° = 0 =
; 𝑚1 = 𝑚2
1 + 𝑚1 𝑚2
Two straight lines having same 𝑚 will be parallel.
69. To find the condition that two straight lines may be perpendicular.
𝑚1 − 𝑚2
tan 90° = ∞ =
; 1 + 𝑚1 ∙ 𝑚2 = 0; 𝑚1 ∙ 𝑚2 = −1
1 + 𝑚1 𝑚2
The straight line 𝑦 = 𝑚1 𝑥 + 𝑐1 is therefore perpendicular to
𝑦 = 𝑚2 𝑥 + 𝑐2 , if 𝑚1 = −
1
𝑚2
.
Questions (6)
Find the angles between the pairs of straight lines
𝑥 − 𝑦 3 = 5 𝑎𝑛𝑑 3𝑥 + 𝑦 = 7.
𝑥 − 4𝑦 = 3 𝑎𝑛𝑑 6𝑥 − 𝑦 = 11.
𝑦 = 3𝑥 + 7 𝑎𝑛𝑑 3𝑦 − 𝑥 = 8
𝑦 = 2 − 3 𝑥 + 5 𝑎𝑛𝑑 𝑦 = 2 + 3 𝑥 − 7.
Find the tangent of the angle between the lines whose intercepts
on the axes are respectively 𝑎, −𝑏 𝑎𝑛𝑑 𝑏, −𝑎.
6. Prove that the points 2, −1 , 0, 2 , 2, 3 , 𝑎𝑛𝑑 4, 0 are
the coordinates of the angular points of a parallelogram and find
the angle between its diagonal.
1.
2.
3.
4.
5.
Questions (4)
Find the equation to the straight line
1. passing through the point 2, 3 and perpendicular to the
straight line 4𝑥 − 3𝑦 = 10.
2. passing through the point −6, 10 and perpendicular to the
straight line 7𝑥 + 8𝑦 = 5.
3. passing through the point 2, −3 and perpendicular to the
straight line joining the points 5, 7 𝑎𝑛𝑑 −6, 3 .
4. passing through the point −4, −3 and perpendicular to the
straight line joining the points 1, 3 𝑎𝑛𝑑 2, 7 .
POINT OF INTERSECTION
Find the coordinates of the points of intersection of the straight lines
whose equations are
1. 2𝑥 − 3𝑦 + 5 = 0 𝑎𝑛𝑑 7𝑥 + 4𝑦 = 3.
2.
𝑥
𝑎
𝑦
𝑥
𝑏
𝑏
+ = 1 𝑎𝑛𝑑
𝑦
+ = 1.
𝑎
3. 2𝑥 − 3𝑦 = 1 𝑎𝑛𝑑 5𝑦 − 𝑥 = 3, and the angle at which they cut
one another.
4. 3𝑥 + 𝑦 + 12 = 0 𝑎𝑛𝑑 𝑥 + 2𝑦 − 1 = 0 and the angle at which
they cut one another.
5. Prove that the following sets of three lines meet in a point.
[1].
2𝑥 − 3𝑦 = 7, 3𝑥 − 4𝑦 = 13, 𝑎𝑛𝑑 8𝑥 − 11𝑦 = 33
[2].
3𝑥 + 4𝑦 + 6 = 0, 6𝑥 + 5𝑦 = −9, 𝑎𝑛𝑑 3𝑥 + 3𝑦 = −5.
The circle: Def. A circle is the locus of a point which moves so that its
distance from a fixed point, called the centre, is equal to a given
distance. The given distance is called the radius of the circle.
Y
P
O
O
M
M
X
139. To find the equation to a circle, having its centre at the origin.
Let O be the centre of the circle and let 𝑎 be its radius.
Let OX and OY be the axes of coordinates. Let P be any point on the
circumference of the circle, and its coordinates be 𝑥 𝑎𝑛𝑑 𝑦.
Draw PM perpendicular to OX and join OP. Then
𝑂𝑀2 + 𝑀𝑃2 = 𝑎2
i.e.
𝑥 2 + 𝑦 2 = 𝑎2
This being the relation which hold between the coordinates of any
point on the circumference is the required equation of the circle.
140. To find the equation to a circle
referred to any rectangular axes.
Y
P
C
O
L
M
N
X
Let OX and OY be the two rectangular axes.
Let C be the centre of the circle and 𝑎 its radius.
Take any point P on the circumference and draw perpendicular CM and
PN upon OX; Let P be the point 𝑥, 𝑦 .
Let the coordinates of C be ℎ 𝑎𝑛𝑑 𝑘; these are supposed to be known.
We have 𝐶𝐿 = 𝑀𝑁 = 𝑂𝑁 − 𝑂𝑀 = 𝑥 − ℎ,
And 𝐿𝑃 = 𝑁𝑃 − 𝑁𝐿 = 𝑁𝑃 − 𝑀𝐶 = 𝑦 − 𝑘.
Hence, since 𝐶𝐿2 + 𝐿𝑃2 = 𝐶𝑃2 ,
We have 𝑥 − ℎ 2 + 𝑦 − 𝑘
2
This is the required equation.
= 𝑎2
Example
The equation to the circle, whose centre is the point −3, 4 and
whose radius is 7, is
2
2
𝑥+3 + 𝑦−4 =7
i.e.
2
2
2
𝑥 + 𝑦 + 6𝑥 − 8𝑦 = 24.
Questions (7)
Find the equation to the circle
1. Whose radius is 3 and whose centre is −1, 2 .
2. Whose radius is 10 and whose centre is −5, −6 .
3. Whose radius is 𝑎 + 𝑏 and whose centre is 𝑎, −𝑏 .
Find the coordinates of the centers and the radii of the circles whose
equations are
1. 𝑥 2 + 𝑦 2 − 4𝑥 − 8𝑦 = 41
2. 3𝑥 2 + 3𝑦 2 − 5𝑥 − 6𝑦 + 4 = 0
Find the equations to the circles which pass through the points
1. 1, 2 , 3, −4 , 𝑎𝑛𝑑 5, −6
2. 1, 1 , 2, −1 , 𝑎𝑛𝑑 3, 2
Acknowledgment
The author acknowledges all the websites
which helped a lot in preparing the slides
presented here meant for B.Sc.(Ag) students.
D C Agrawal
dca_bhu@yahoo.com
Note: There are couple of typographical errors;
you are supposed to find it out.