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COORDINATES GEOMETRY
Name
........................................................................................
Coordinate Geometry
CHAPTER 6 : COORDINATE GEOMETRY
A. Distance between two points.
Given point A (x1, y1) and B (x2, y2) on a Cartesian plane.
Formula used: AB = 2
12
2
12 )()( yyxx
Exercises:
1. Find the distance between the following points:
Q Points Given Distance
1. a) A (2,5) and B (6, 8) AB = c 22 )58()26( =
22 )3()4( = 5
b) C (-5, -6) and D (1, -2)
c) E (3, -2) and F (8,10)
d) G (-5,-2) and H (1,6)
2. Find the value of p and the value of q for each of the following
Q Information Given Solution
2. a) R (p, 3) and S (-4, -1)
and RS = 5
√[p-(-4)]2
+ [3-(-1)]2
=5
p2 + 8p+ 16 +16 = 25
p2 + 8p + 7 = 0
(p+1) (p+7) = 0
p = -1, p = -7
b) T (2, 1) and V (9, q)
and TV = 10
y
x
B (x2, y2)
A (x1, y1)
0
Coordinate Geometry
3.
a) P (12, 2) and Q (0, y), PQ
= 13 Find point Q.
√(12 – p)2 + (2 – y)
2 = 13
144 + 4 – 4y + y2 = 169
y2 – 4y -21 = 0
(y – 7)(y + 3) = 0
y = 7 or y = -3
Accept y = 7
Therefore, Q (0, 7)
b) R (-12, -5 ) and S ( 0, y )
and RS = 15. Find point S.
Activity:
1.
In the diagram, ABC is a triangle. Find the distance of each of the following.
a) AB
b) AC
c) BC
2. Find the possible values of a if the distance between points A (-1, -3) and B (a, 2) is √50
units.
A (1, 2)
B (6, 1)
C (2, -2)
x
y
P (12, 2)
Q (0, y)
R (-12, -5)
S
0
0
0
Coordinate Geometry
3. The distance between point A (-k, 1) and the origin is the same as the distance between point
B (k, 1) and point C (3, 5). Find the value of k.
4. Given point P (h, k) is equidistant from points A (2, 5) and B (-2, 4). Show that 2k + 8h =9.
5. The coordinates of points A, B and C are (1, 2), (-5, 1) and (p, 3) respectively.
a) Find the distance of AB.
b) Given that AB = AC, find the two possible values of p.
Answers:
1. a) √26 b) √17
2. -6, 4
3. 4
5. a) 6.08
b) p = 7, -5.
B. Division of a line segment.
B.1 Midpoint of two given points
B.2 Coordinates of a point that divides a line according to a ratio m: n.
________________________________________________________________________
B.1 Midpoint of two given points.
A (x1, y1)
B (x2, y2)
P (x, y)
Formula:
Midpoint of AB
P (x, y) =
2
yy,
2
xx 2121
Coordinate Geometry
Exercises:
1. Find the midpoint of the two given points :
Q. Points Given Solution
1. a) A (6, -8) and B (-4, 4)
2
48,
2
)4(6
= (1, -2)
b) C (4, -5) and D (6, 13)
c) E (-4,6) and F (2, -10)
2. The coordinates of points R and S are (-4,3) and (x,y) respectively. Point P(-1,2) is the
midpoint of RS. Find the values of x and y.
Q Points Given Solutions
2. a) R (-4, 3) and S (x, y). Midpoint P
is (-1,2)
2
)3(,
2
)4( yx
= (-1, 2)
2
)4( x = -1
x = 2
2
)3( y = 2
y = 1
Therefore, S (2, 1)
3. The coordinates of points G and H are (5,-2) and (x,y) respectively. Point Q is the midpoint
of GH. Find the value of x and y.
Q Points Given Solutions
3 G (5,-2) and H(x,y). Midpoint Q is
(1,8)
Coordinate Geometry
B.2 Coordinates of a point that divides a line according to a ratio m: n.
Exercises:
Q. Points Given Ratio (m:n) P (x, y)
1. a) A (-7, 5)
B (2, -1)
1 : 2
3,4
21
1152,
21
2172
b) A (2, 1)
B (8, 4)
2 : 1
c) A (3, 5)
B (-12, 0)
3 : 2
A (x1, y1)
Formula:
P (x, y) = [(mx2 + nx1), (my2 + ny1)]
m + n m+n
B (x2, y2)
P (x, y)
m
n
Coordinate Geometry
Q
Questions
Solutions
2. The points A and B have
coordinates (-7,2) and (x,y).
Point P(1,6) divides the line
AB internally in the ratio
4:1. Find the coordinates of
B.
7,3
7,65
24
3,15
74
6,1
5
24,
5
74
41
214,
41
714
B
yy
xx
yx
yx
3 P(-3,-2) is a point dividing
the line segment joining two
points (9,1) and S9-6,-3)
intenally in the ratio of m:n.
Find m:n.
1:3:
1
3
3
9
93
3636
3366
366
)2,3(
)1()3(,
)6()6(
nm
n
m
nm
nnmm
nmnm
nm
nm
nm
nm
nm
nm
Activity
1. The coordinates of points P and Q are (4, k) and (h, 5) respectively. Point R (-1, 2) is
the midpoint of PQ. Find the values of h and k.
2. If M is a point on the straight line AB such that AM = BM, find the coordinates of point
B if the coordinates of points A and M are (-3, 1) and (1, 4) respectively.
Coordinate Geometry
3.
In the diagram, points P and Q are the midpoint of the line segments AB and PC
respectively. Find the coordinates of point C.
4. The points R and S have coordinates (-5, 1) and (7, -3) respectively. Point P lies on the
line segment RS such that RP = 3 PS. Find the coordinates of point P.
5. T (a, 1) is a point dividing the line segment joining two points U (4, 3) and V (-5, 0)
internally in the ratio m: n. Find
a) m: n
b) the value of a.
Answers:
1. h = -6, k = -1
2. B (5, 7)
3. C (7, 4)
4. P (4, -2)
5. a) 2: 1
c) a = -2
C A (-6, 1)
P B (4, -5)
Q (3, 1)
Coordinate Geometry
)()(2
1
2
1
5443322114433221
54321
54321
xyxyxyxyyxyxyxyx
yyyyy
xxxxx
C. Area of Polygons
C.1 Area of triangle
Note:
1. Anticlockwise direction – positive.
2. Clockwise direction – negative
3. Area zero: A, B, C are collinear.
C.2 Area of a quadrilateral.
Area of ABCD
Reminder: Points are arranged in anticlockwise/ clockwise direction.
Suggestion: Sketch the points first before using the formula.
A (x1, y1)
B (x2, y2)
C (x3, y3)
0
Coordinate Geometry
Exercises:
Find the area for each of the following.
Q. Points given Area
a) A (1, 2), B (5, -3), C (-1, -5)
216
382
1
8302
1
)5(310)2()25(32
1
2532
1151
2
1
unit
b) D (-4, 2), E (-3, 6), F (6, 4) and G (2, 1)
230
34262
1
483664612242
1
21462
42634
2
1
unit
c) P (-5, 3), Q (0, 20, R (-1, 7)
d) A (6, 5), B (-1, 4), C (-2, 1), D (5, 2)
F
G
E
D
x
y
0
Coordinate Geometry
Activity
1. The vertices of PQR are P (5, 4), Q (-4, 1) and R (k, 0). Given that the area of PQR
is 15 unit2, find the possible values of k.
2. Show that points A (-8, 1), B (-4, 5) and C (-2, 7) are collinear.
3. The vertices of quadrilateral PQRS are P (5, 2), Q (a, 2a), R (4, 7) and S (7, 3). Given that
the area of quadrilateral PQRS is 12 unit2, find the possible values of a.
4. Given T (-2, 7), U (2, 4), V (3, 7) and W are the vertices of a rectangle PQRS. Find W.
Answers:
1. k =3, -17
2. a = 8 6/7 or 2
3. a) W (-1, 10)
b) 15 unit2
Coordinate Geometry
D. Equation of a straight line
D.1 x-intercept and y-intercept
D.2 Gradient of a straight line that passes through two points
D.2.1
y-intercept = b
x-intercept = a
(a, 0)
(0, b)
m = gradient
= b – 0
0 – a
= - b
a
A (x1, y1)
B (x2, y2)
Gradient of AB
mAB = y2 – y1
x2 – x1
x
y
x
y
0
0
Coordinate Geometry
D.2.2 Types of gradient
Positive gradient
Negative gradient
Gradient = 0
Gradient = undefined
Coordinate Geometry
Exercises
1. Find the x-intercept , y-intercept and the gradient for each of the following
Q. Graphs x-intercept y-intercept Gradient
1. a)
b)
c)
2. Find the gradient for each of the following.
Q. Points Given Gradient
2. a) (1, 3) and (4, 9)
b) (-1, 2) and (1, -8)
c) (2, -4) and (4, 8)
(0, 2)
(6, 0)
(0, 5)
(-3, 0)
(0, -2)
(-4, 0)
y
y
y
x
x
x 0
0
0
Coordinate Geometry
-3
-3
2
0
0
0
Activity:
1. Find the value of h if the straight line joining the points (2h, -3) and (-2, -h + 2) has a
gradient of 2.
2. Find the value of k if the points A (-3, 3k), B (0, k) and C (3, -2) are collinear.
3. Find the gradient of the following straight lines.
a) b)
4. The diagram shows a straight line which has a gradient of ½. Find the coordinates of
point A.
Answers:
1. h = -3
2. k = 2
3. a) 2
3
b) 3
5
4. A (4, 0)
5
-2
A
y y
y
x
x
x
Coordinate Geometry
D.3 Equation of a straight line
Given Equations
Gradient = m
Point = (x, y)
1y y = m 1( )x x
Points : (x1, y1)
and : (x2, y2)
1
1
y y
x x
= 2 1
2 1
y y
x x
x-intercept = a
y-intercept = b
x + y = 1
a b
Exercises:
1. Find the equation of a straight line for each of the following
Q. Given Equation
1. a) m = 2
point : (3, 1)
y – 1 = 2 ( x – 3 )
y = 2 x – 6 + 1
y = 2 x - 5
b) Points (-3, 1) and (-1, -3)
c) x-intercept = 4
y-intercept = -2
Coordinate Geometry
Activity:
1. Given the points P (5, 7), Q (8, 0) and R (-4, 8), find the equation of a straight line that
passes through P and the midpoint of QR.
2. A straight line has a gradient of 4
h and passes through a point (0, 4a).
a) Find the equation of the straight line.
b) If the straight line passes through point (-4, 3), find the value of h.
3. Find the equation of a straight line given the x-intercept and the y-intercept are 3 and -4
respectively.
Answers:
1. y = x + 2
2. a) 4y = h x + 16h b) h= 1
3. 3
x –
4
y = 1
Coordinate Geometry
E. Gradient and intercepts of a straight line.
Forms of equation Equation
1. Gradient form
y = mx + c
2. Intercept form
x
a +
y
b =1
3. General form
ax + by + c = 0
Examples:
1. Given the equation 3y + 7x – 9 = 0, find each of the following:
a) gradient of the line
b) y-intercept
c) x-intercept
Solution:
a) 3y + 7x – 9 = 0
3y = -7x + 9
y = 7
3
x + 3 - gradient form
Gradient = 7
3
b) x = 0, y-intercept = 3
c) y=0, 7
3
x + 3 = 0
7
3
x = 3
x = 9
7
Therefore, x-intercept = 9
7
Exercise
Question 1 (see activity)
Coordinate Geometry
F. Equation of a straight line in general form
2. Express the following equations in the general form.
General Form
a) y = 2x + 7 y – 2x – 7 = 0
b) y = 1
3
x + 3
y = 1
3
x + 3
3y + x – 9 = 0
c) 3
x +
5
y = 1
5x + 3y – 15 = 0
Exercise : Question 2 & 3 (refer Activity)
Coordinate Geometry
Activity:
1. Find the gradient and the y-intercept of each of the following equations.
a) 3x – 5y – 15 = 0
b) 2x + 3y = 30
2. Given that the gradients of the straight lines 6
x +
y
a = 1 and 3y + x – 4 = 0 are the
same, find the value of a.
3. Express the following equations in the general form
a) y = 3x + 4
b) y = - 1
2x + 5
c) 2y – 3 = 1
3 x
d) 2
x+
3
y = 1
Answers:
1. a) 3
5, y-intercept = -
b) m = 2
3 , y-intercept = 10.
2. a = 2
3. a) 3x – y + 4 = 0
b) x + 2y – 10 = 0
c) x – 6y + 9 = 0
d) 3x + 2y – 6 = 0
Coordinate Geometry
G. Point of intersection of two lines
Note: Solve two simultaneous equations
to get point of intersection.
Example:
1. Find the point of the straight lines y = -2x + 1 and y = ½ x + 6
Solution:
y = -2 x + 1…….. (1)
y = 1
2 x + 6…….. (2)
(1) = (2), -2x + 1 = 1
2x + 6
½ x + 2x = 1 -6
x + 4x = -5
5x = -5
x = -1 in (1)
y = -2 (-1) + 1
= 3
Therefore, point of intersection is (-1, 3)
Problems involving points of intersection
Example
2. A straight line passes through points (1, 5) and (6, 0) and intersects with the straight
line 2y = 5x – 16 at point P. Find the coordinates of point P.
Points (1, 5) and (6, 0)
(y – 5) = (0 – 5) = -5 = -1
(x – 1) (6 – 1) 5
y – 5 = - (x – 1) = -x + 1
y = - x + 6 ……. (1)
2y = 5x – 16 ………. (2)
(1) in (2), 2 (-x + 6) = 5x – 16
-2x + 12 = 5x – 16
Simultaneous equations
Simultaneous equations
Coordinate Geometry
7x = 28
x = 4 in (1)
y = -4 + 6 = 2
Therefore, P (4, 2)
Activity
iuPO
1. A straight line intersects both the axes at 2 and intersects with the straight line y = 3x –
2 at point A. Find the coordinates of point A.
2.
The diagram shows straight lines AB and BC with the equations 2y = x + 10
and y = 7x – 2 respectively. AB intersects the y-axis at A and BC intersects the
x-axis at C. Find
a) The coordinates of points A, B and C.
b) The equation of AC in the intercept form.
3. A straight line passes through a point (5, 1) and the x-intercept is 10. If the straight line
intersects the y-axis at point R, find
a) The equation of the straight line.
b) The coordinates of point R.
4. Two straight line y
b –
2
x = 1 and ky = -x + 12 intersect the y-axis at the same point.
Find
a) The value of k.
b) The gradient of the straight line ky = -x + 12.
y = 7x - 21
2y = x + 10
A
C
B
y
x 0
Coordinate Geometry
Answers:
1. A (1, 1)
2. a) A (0, 5), B ( 4, 7) and C ( 3, 0)
b) 3
x +
5
y = 1
3. a) 5y = -x + 10
b) R (0, 2)
4. a) k = 2.
b) - 1
2
H. Parallel and Perpendicular Lines
Parallel Lines Perpendicular Lines
m1 = m2
m1 m2 = -1
Example:
1.Given points P (-4, 8), Q (-1, 5), R (-7, 5)
and S (-1, -1), determine whether PQ and
RS are parallel lines.
Solution:
mPQ = (8-5)/ [-4-(-1)] = -1
mRS = [5 – (-1)] / [-7 – (-1)] = -1
Therefore, m PQ = m RS
PQ and RS are parallel
Example:
Given points A (-8, -1), B (2, 4), C (-3, 7)
and D (2, -3), determine whether straight
lines AB and CD are perpendicular to each
other.
Solution:
mAB = (-1 – 4)/ (-8 – 2) = 1
2
mCD = [7- (-3)]/ (-3 – 2) = -2
mAB mCD = 1
2 (-2) = -1
AB CD
m1 m2
m2
m1
Coordinate Geometry
Exercises
2. Determine whether the following pairs of straight lines are parallel or perpendicular.
Equations Value of m1 and m2 Parallel Perpendicular
a) y = 5x + 5
y = 5x - 4
m1 = 5
m2 = 5
Yes -
b) 5y = x + 3
y = -5x - 3
m1 = 1/5
m2 = -5
-
Yes
c) 2y = 4x + 3
y = 2x - 3
d) x/5 + y/2 = 1
2y = 5x + 1
I. 1. Equation of a straight line that passes through a fixed point and
parallel to a given line.
Given
Use y - y 1 = m 1 ( x - x 1 )
y = m 1 x + c
m1 = m2
P 1 1( , )x y
Coordinate Geometry
Example
Find the equation of a straight line that passes through point (2, 3) and is parallel to the
straight line 2y – 6x = 9.
Solution:
2y – 6x =9
2y = 6x + 9
y = 3x + 9
2
Therefore, m= 3
Hence, y - y1 = m(x – x1)
y – 3 = 3 (x – 2)
y – 3 = 3x – 6
y = 3x – 3
Exercise : Question number 1 ( refer Activity )
Coordinate Geometry
I. 2. Equation of a straight line that passes through a fixed point and is
perpendicular to a given line.
Example:
Find the equation of a straight line that passes through point (1, -5) and is
perpendicular to the straight line 3y – 2x = 4.
Solution:
3y – 2x = 4
3y = 2x + 4
y = 2/3 x + 4/3
m1 = 2/3 m2 = -3/2
Hence, y – (-5) = -3/2 (x – 1)
2y + 10 = -3x + 3
2y = -3x – 7.
Exercise : Question 2,3,4 and 5 ( refer Activity )
Given:
Find: The equation of this line
Use: y – y1 = - 1
1
m (x – x1)
y1 = m1x + c
Gradient =-1
1
m
Tips:
Perpendicular bisector
Q(x1, y1) use
m1 m2 = -1
Coordinate Geometry
Activity
1. Find the equation of a straight line that passes through point (1, -2) and is parallel to the
straight line 2y – 3x + 4 = 0.
2. Find the equation of a straight line that passes through passes through point (3, 4) and is
perpendicular to the straight line 2y + x + 3 = 0.
3. The coordinates of points A and b are (-3, -2) and (5, 4) respectively. Find the equation
of the perpendicular bisector of AB.
4. Given point P (-3, 11) and Q (1, 3), find the equation of the straight line that passes
through the midpoint of PQ and is perpendicular to the straight line 2y = x + 7.
5. A straight line x + ky = 10 is perpendicular to the straight line y – 4x = 28. Find
a) The value of k.
b) The point of intersection of the two lines.
Answers
1. 2y – 3x + 7 = 0.
2. y – 2x + 2 = 0.
3. 3y + 4x – 7 = 0.
4. y + 2x = 5.
5. a) k = 4
b) (-6, 4)
Coordinate Geometry
J. Equation of a Locus
A. Locus of a point that moves in such a way that its distance from a fixed point is a
constant.
B. Locus of a point that moves in such a way that the ratio of its distance from two fixed
Points is a constant.
Tips: Given - points, C and D such that CP : PD = 1 : 3.
- PC: CD = 1: 3
- CP
CD =
1
3
- 3 PC = CD
-
Use:
2 2
1 1(x- x ) (y - y ) k
Locus P
A (x1, y1)
P (x, y)
A (x1, y1)
P (x, y)
B (x2, y2)
Locus P
Use: 2222 y2) -(y x2)-(x y1) -(y x1)-(x
k unit
C
P
D
1
3
0
Coordinate Geometry
Example :
1. A point P moves such that its distance from a fixed point A (2, 1) is 3 units. Find the
equation of the locus of P.
Solution:
Let P (x, y), PA = 3.
3 1) -(y 2) -(x 22
(x – 2)2 + (y – 1)
2 = 9
x2 – 4x + 4 + y
2 – 2y – 4 = 0
Therefore, x2 + y
2 – 4x – 2y – 4 = 0.
Exercise : Question number 1 ( refer next page )
Example:
2. A point P moves such that it is equidistant from points A (2, -1) and B (3, 2). Find the
equation of the locus of P.
Solution:
Let P (x, y), PA = PB.
22222312-9x yxy
x2 – 4x -4 + y
2 + 2y + 1 = x
2 – 6x + 9 + y
2 – 4y + 4
2x + 6y = 8
There fore, x + 3y = 4
Exercise: Questions number 2,3,4 and 5 ( refer next page )
Coordinate Geometry
Activity
1. Find the equation of the locus of a point P which moves such that its distance from each
fixed points as follows,
a) 3 units from A (-2, 1)
b) 5 units from B (-3, -1)
2. A point P moves such that it is equidistant from points A (3, 2) and B (2, 1). Find the
equation of the locus of P.
3. A point P moves such that its distance from two fixed points R (-1, 4) and S (1, 3) is in
the ratio 1:2. Find the equation of the locus P.
4. A point P moves such that its distance from points A (0, -2) and B (0, 3) is in the ratio
of 2:3.
a) Show that the equation of the locus P is x2 + y
2 + 12y = 0.
b) Show that point C (3, -3) does not lie on the locus of P.
5. The distance of a moving point P (x, y) from point R (3, 0) is twice its distance from
point S (0, 5).
a) Find the equation of the locus of P.
b) Determine whether the locus of P intersects the x-axis.
Answers
1. a) x2 + y
2 + 4x – 2y – 4 = 0
b) x2 + y
2 + 6x + 2y – 15 = 0
2. x + y = 4
3. 3x2 + 3y
2 + 10x – 26y + 58 = 0
5. a) 3x2 + 3y
2 + 6x – 40y + 91 = 0
b) Does not intersect.
Coordinate Geometry
Advance Activity (Cloned SPM Questions)
Paper 1
1. Find the equation of the straight line that passes through the point (2, 5) and is
perpendicular to the straight line 2x – y + 7 = 0 [3M]
2. A straight line 3x – 4y = 24 meets the x-axis at point A. it meets the y-axis at point B.
a) Find the coordinates of A and B.
b) Calculate the area of triangle AOB, where O is the origin. [3M]
3. A point P (x, y) moves such that its distance from a fixed point Q (2, 4) is 5 units. Find
the equation of the locus of P. [3M]
4. If the distance between the points P (2, k) and Q (1, 2) is √5 units, find the possible
values of k. [3M]
5. A point P (x, y) moves such that its distance from a fixed point A (1, 2) is two times its
distance from another fixed point B (-1, 1). Find the equation of the locus of P.
[4M]
6.
Diagram shows the positions of three points A (-3, 5), B (7, 0) and C (-1, 6). It is
given that D is a point which divides the line AB in the ratio 3:2. Find
a) The coordinates of D.
b) The distance of CD. [4M]
y
x
D (x, y)
B (7, 0)
A (-3, 5)
C (-1, 6)
0
Coordinate Geometry
Advance Activity (Cloned SPM Questions)
Paper 2
7.
A straight line 2x + 3y – 12 = 0 meets the x-axis and the y-axis at the points P and Q
respectively. A straight line L, which is perpendicular to PQ, passes through the
midpoint of PQ and meets the y-axis at a point R.
a) Find the coordinates of P and Q.
b) Find the equation of the straight line L.
c) Calculate the area of triangle PQR. [10 M]
8.
y
x
R
Q
P
L
y
x
E
B (0, k)
D
A 2x + 3y + 6 = 0
C
0
0
Coordinate Geometry
Diagram shows two perpendicular lines ABC and DBE. Find
a) The value of k.
b) The equation of the line DBE.
c) The coordinates of the point of the intersection of the line DBE with the x-
axis. [10 M]
9.
Diagram shows a triangle POQ where O is the origin. Point R lies on the straight line
PQ.
a) Calculate the area, in unit2, of triangle POQ.
b) Given that PR: RQ = 2: 1, find the coordinates of R.
c) A point S moves such that its distance from point P is always 4 units
i) Find the equation of the locus of S.
ii) Hence, determine whether or not this locus intersects the y-axis.
[10 M]
y
x 0
P (1, 5)
R
Q (7, 1)
Coordinate Geometry
Answers (Advance Activity)
1. x + y = 12
2. a) A (3, 0)
B (0, -6)
b) 24 unit2
3. 3x2 + 3y
2 + 10x – 4y +3 = 0
4. x2 + y
2 – 4x -8y -5 = 0
5. k = 4 or 0.
6. a) (3, 2)
b) 5.657 units.
7. a) P (6, 0) , Q (0, 4)
b) y = 3/2 x – 5/2
c) 19.5 unit2.
8. a) k = -2
b) y = 3/2 x – 2
c) (4/3, 0)
9. a) 17 unit2
b) R (5, 7
3)
c) i) x2 + y
2 – 2x – 10y +10 = 0
ii) Yes