Unit 6.2 : Division Division of a Line Segment Segment 6.2.1 To find the mid-point of Two Given Points. Formula : Eg.
Midpoint M =
P(3, 2) an and Q(5, 7)
E1
P(-4, P(-4, 6) and Q(8, Q(8, 0)
3 5 2 7 , 2 2
Midpoint, Midpoint, M = = E2
x1 x 2 y1 y 2 , 2 2
(4 ,
9
)
2
(2, 3)
P(6, P(6, 3) and and Q(2, Q(2, -1)
E3
P(0,P(0,-1), 1), and Q(-1, Q(-1, -5)
(4, 1)
(- ½ , -3)
6.2.2 Division of a Line Segment Q divides the line segment PR in the ratio PQ : QR = m : n. P( x, y), R( x, y)
m P(x1, y 1)
Q (x,y) =
n
n
Q(x, y)
R(x2, y2)
m
R(x2, y2)
●
Q(x, y)
nx1 mx2 ny1 my 2 , m n m n
P(x1, y 1)
(NOTE : Students Students are strongly advised advised to sketch a line segment segment before applying applying the formula) formula)
Eg1. The point P internally divides the line segment E1. The point point P intern internall ally y divide dividess the line line segmen segmentt joining the point M(3,7) and N(6,2) in the ratio 2 : 1. joining joining the point M (4,5) and N(-8,-5) N(-8,-5) in the ratio Find the coordinates of point P. 1 : 3. Find the coordinates of point P.
1 2
N(6, N(6, 2)
●
P(x, y)
M(3, 7)
1(3) 2(6) 1(7) 2(2) , 2 1 2 1 15 11 = , 3 3 11 = 5, 3
P=
6. Coordinate Geometry
5 1, 2
2
More Exercise Exercise : The Ratio Theorem Theorem (NOTE : Students Students are strongly advised advised to sketch a line segment segment before applying applying the formula) formula)
E1. R divi divides des PQ in the ratio ratio 2 : 1. Find Find the the coordi coordinat nates es of R if
E2. P divi divides des AB in the ratio ratio 3 : 2. Find Find the the coordi coordinat nates es of P if
(a) P(1, 2) and Q( -5, -5, 11)
(c) A(2, -3) -3) and B( -8, -8, 7)
(b) P(-4, P(-4, 7) and Q(8, -5) -5)
(d) A(-7, 5) 5) and B(8, -5) -5)
(a) (-3, 8)
(b) (4 , -1)
(a) (-4, -3)
E3. M is a point that lies on the straight line RS such that 3RM = MS. If the coordinates of the points R and and S are are (4,5 (4,5)) and and (-8, (-8,-5 -5)) resp respec ectiv tivel ely, y, find find the the coordinates of point M.
(b) (2 , -1)
E4. P is a point that that lies on the straigh straightt line TU such such that 3TP = 2PU. If the coordinates of the points T and U are are (-9, (-9,7) 7) and and (1,(1,-3) 3) resp respec ecti tiv vely ely, fin find the the coordinates of point P.
3RM = MS
RM MS
=
1 3
,
RM : MS = 1 : 3
Ans : 1, 5
2
(-5, 3)
E5. The points P(3, p), B(-1, 2) and C(9,7) lie on a straight line. If P divides BC internally in the ratio m : n , find find (a) (a) m : n , (b) (b) the the value alue of p.
E6. R( x, y) , divides divides the points points P(2k, P(2k, – k) k) and Q(2x, 4y) in the the ratio ratio 3 : 5. Expr Expres esss x in terms of y.
(a) 2 : 3
(Ans (Ans : x = 4y) 4y)
6. Coordinate Geometry
(b) p = 4
3
Unit 6.3 6.3 To Find Find Areas Areas of of Polygons Polygons Area Area of a poly polygo gon n
=
1
x1
x2
x3
...
x1
2
y1
y2
y3
...
y1
Note : The area area found found will will be positive positive if the coordin coordinates ates of of the points points are are written written in the anti-c anti-clockw lockwise ise order, and negative if they are written in the clock-wise order. Example 1 : Calculate the area of a triangle given : E1. P(0, 1), Q(1, 3) and R(2,5)
Area of ∆ PQR =
1. P(2,3), Q(5,6) and R(-4,4) Area of ∆ PQR =
1 0 1 2 0 2 1 3 5 1
=
= 12 12 unit unit
2
17 unit2 2
2. The coordinates coordinates of the triangle triangle ABC are (5, 10), (2,1) and (8, k) respec respectiv tively ely.. Find Find the possible possible values values of k, 2 given that the area of triangle ABC is 24 units .
3. The coordinate coordinatess of the triangle triangle RST are (4, 3), (-1, 1) and (t, -3) respectively. Find the possible values of t , given that 2 the area of triangle RST is 11 units .
k = 3 , 35
t = 0 , -22
ii) Area of a quadrilateral =
1 x1 x 2 x3 x 4 x1
2 y1 y 2 y3 y 4 y1
2. P(2, -1), Q(3,3), Q(3,3), R(-1, 5) and S(-4, -1).
1. P(1,5), Q(4,7), R(6,6) and S(3,1). Area Area of PQRS PQRS =
= 8 unit
2
[27]
Note : If the area is zero, then the points are collinear. 1. Given that that the points P(5, 7), Q(4, 3) and R(-5, k) are collinear, find the value of k.
2. Show that the points points K(4, 8), L(2, 2) and M(1, -1) are collinear.
k= 33 6. Coordinate Geometry
4
Unit 6.4 : Equations of Straight Lines The Equation of a Straight line may be expressed in the following forms: i)
The general form :
ax + by + c = 0
ii) ii)
The The gradient form form :
y = mx + c ;
iii) iii)
The The inte interc rcep eptt form form :
x a
+
y b
=1,
m = gradient
,
c = y-in y-inttercept
a = x-inte x-interce rcept pt , b = y-inter y-intercep cept t
Eg. Find the equation of a straight line that passes
a) If given given the the gradien gradientt and one point: point:
through the point (2,-3) and has a gradient of
y y1 = m( x x1 )
1 4
.
y y1 = m( x x1 )
● P(x1 y1 )
y (3)
1 4
( x 2)
Gradient = m
4 y x 14 E1. Find the equation equation of a straight line that passes passes through the point (5,2) and has a gradient of -2.
E2. Find the equation of a straight line that passes through the point (-8,3) and has a gradient of
y = -2x + 12
b) If two two points points are given : Note : You may find the gradient first, then use eithe eitherr (a) (a) y = mx + c Or (b) y – y1 = m( x – x 1) Or (c)
y y1 x x1
=
3 4
.
4y = 3x + 36
Eg. Find the equation of a straight line that passes through the points (-3, -4) and (-5,6) y (4)
= 6 (4) x (3) 5 (3)
y 2 y1 x 2 x1 y 5 x 19
E1. Find the equation equation of a straight line that passes passes through the points (2, -1) and (3,0)
E2. Find the equation of a straight line that passes through the points (-4,3) and (2,-5)
y=x-3
6. Coordinate Geometry
4x + 3y +7 = 0
5
x-intercept and the y-intercept y-intercept are given: c ) The x-intercept m =-
y int ercept x int errcept
E1. The x-intercept and the y-intercept of the straight line PQ are 4 and -8 respectively. Find the gradient and the equation of PQ. m PQ = –
Equation Equation of Straight Straight Line is :
x a
+
y b
8 = – 4 = 2
=1
Note : Sketch a diagram to help you !
At At th thee x-a x-axis xis,, At At th thee yy-axi axis, s,
y int ercept int x errcept
Equation :
y=0 x=0
x
4
+
y
O
y
8
=1
x
4
-8
y 2 x 8
E2. The x-in x-inte terc rcep eptt and and the the y-in y-inte terc rcep eptt of the the straight line PQ are -6 and 3 respectively. Find the gradient and the equation of PQ.
E3. The x-intercept x-intercept of a straight line AB is -5 and its gradient gradient is -3. Find the y-intercept of the straight line AB and the equation of AB.
3x + 5y +15 = 0
2y = x+6 Extra Vitamins for U…… 1. Find the gradient and the equation of AB.
2. The x-intercep x-interceptt of a straight straight line RS is – 2 and its gradien gradientt is 3. Find the y-interce y-intercept pt of the straight straight line RS and the equation of RS.
y
B
6
x
O -2
A
x – 3y = 6
3. Find the equation of KL in the intercept form.
4. Find the equation of the line which connects the origin and the point S (-2, 6).
y
K
y = 3x + 6
3 6
x
O
L
x
6
y
3
1
5. For Q3 above, write down the equation of KL in the general form.
y = – 3x 3x
6. Write down down the equation of the straight straight line which passes passes through through the points P(3, 2) and Q (3, 8).
x + 2y – 6 = 0
6. Coordinate Geometry
[x = 3]
6
Unit 6.5 6.5 Parallel Lines and Perpendicu Perpendicular lar lines lines 6.5. 6.5.1 1
Para Parall llel el line lines, s, m1 = m2
6.5.2 6.5.2
Perpen Perpendic dicula ularr lines, lines, m1 m2 = -1 -1
Unit 6.5.1 Determine whether each of the following pairs of lines are parallel. Eg. y = 3x – 2 and
3x – y = 4
1. y = 2x +5 and
4x + 2y = 5
y = 3 x – 2 , m1 = 3 3x – y = 4 y = 3x 3x – 4 , m2 = 3 Sinc ince m1 = m2 , the two line are parallel . N
2. 3x – 3y = 7 and and 6x + 6y = – 5
3. 2x – 3y 3y = 5 and and 6y = 4x + 9
N
4. x – 3y = 12 and and 6y = 3 + 2x
Y
5.
x
3
y
2
4 and and 8y = 6x - 3
Y
N
Unit 6.5.2 Determine whether each of the following pairs of lines are perpendicular. Eg. Eg. y = 3x – 2 and
x + 3y = 4
y =3x–2 , x + 3y = 4 3y = – x + 4
y
1 3
Sinc Sincee m1 . m2 =
x
1. y = 2x +5 and
4x + 2y = 9
m1 = 3
4
, m2 =
3
3
1 3
1 3
1 ,
The two given lines are perpendicular . 2.
3y = 2x – 2 and
N
2x + 3y = 1
3. x – 3y = 2 and
6x + 2y = 5
N
4.
6y = 2 - 3x and and
x
3
y
6
4
Y
5.
Y
6. Coordinate Geometry
x
3
y
4
1 and and 8y + 6x 6x – 3 = 0
Y
7
6.5.2 Applica Applications tions (m1 .m 2 = – 1) Ex.1 Ex.1 (SPM (SPM 200 2004) 4).. Diagram 1 shows a straight line line PQ with with the the equa equatio tion n
x
2
y
4
Ex.2. Diagram Diagram 2 shows a straight straight line PQ with the
Find the equation 1 . Find
x
6
y
2
ind the the equa equati tion on of the the 1 . Find
equatio equation n of the straig straight ht line line perpen perpendic dicula ularr to PQ stra straig ight ht line line perp perpen endi dicu cula larr to PQ and and pass passin ing g and passing through the point Q. through the point P. y
y Q Q
Diagram 1 P O
Diagram 2 P
x
O
Answer:
x
Answer:
y = ½ x + 4
y = 3x – 18
Ex.3 Ex.3 Diagram Diagram 3 shows shows a straig straight ht line line RS with the the equa equati tion on x + 2y = 6. Find Find the equa equati tion on of the stra straig ight ht line line perp perpen endi dicu cula larr to RS and and passi passing ng through the point S.
Ex.4. Diagram Diagram 4 shows a straight straight line AB with the equa equati tion on 2x – 3y = 6. Fin Find d the equa equati tion on of the the stra straig ight ht line line perp perpen endi dicu cula larr to AB and and pass passin ing g through the point B.
y
y R
O
Diagram 3
B
x
A S O
Diagram 4
x
Answer:
Answer:
y = 2x – 12
6. Coordinate Geometry
2y = 3x – 9
8
6.5.2 Applica Applications tions (m1 .m 2 = – 1) – more exer exercise cisess Ex.5 Ex.5 Diagra Diagram m 5 shows a straig straight ht line line PQ with the the Ex.6. Diagram 6 shows a straight line AB with the equa equatio tion n 4x + 3y = 12. 12. Find the the equatio equation n of the x y equation ind the the equ equatio ation n of the the 1 . Find stra straig ight ht line line perp perpen endi dicu cula larr to RS and and pass passin ing g 4 6 through through the midpoint midpoint of RS. perpendicular bisector of the line AB. y
y R O
Diagram 5 S O
A
x
Answer:
B
x
Diagram 6
Answer:
4x+3y = 8
2x + 3y = 6
Ex.7 Ex.7.. Find Find the the equa equatio tion n of the the stra straig ight ht line line that that Ex.8 Find the the equation equation of the the straight straight line that that passes through the point ( 1, 2) and is perpendicular passes through the point (3, 0) and is perpendicular to the straight line x + 3y +6 = 0. to the straight line 3x – 2y = 12.
y = 3x – 1
Ex.9 Ex.9 Find Find the equatio equation n of the stra straig ight ht line that that passes through the origin O and is perpendicular to the straight line that passes through the points P(1, – 1 ) and Q(-3,7 Q(-3,7). ).
2x+3y = 6
Ex. 10 Find the equation equation of the straight straight line that that passes passes through through the point (-2,4) (-2,4) and is perpendicular to the straight line which passes through the origin O and the point (6, 2).
y = -3x
y = ½ x
6. Coordinate Geometry
9
Unit Unit 6.6 Equatio Equation n of a Locus Locus Note : Students Students MUST be able to find distance distance between two points points [ using Pythagoras Pythagoras Theorem] Theorem]
TASK : To Find the equation equation of the locus locus of the moving moving point P such that that its distances distances of P from the points Q and R are equal. Eg 1. Q(6, Q(6, -5) and and R(1, R(1,9) 9) R(1, 9)
●
Let Let P = (x,y (x,y), ), then then PQ = PR 2 2 ( x 6) ( y (5) =
2 2 ( x 1) ( y 9)
Q(6, -5)
Square both sides to eliminate the square roots.
x 6)
( y 5 = x 1) ( y 9 x 2 12 x 36 y 2 10 y 25 x 2 2 x 1 y 2 18 y 81 10 x 28 y 21 0 2
2
2
E1. Q(2,5) and R(4,2)
2
●
●P(x, y) Locus of P
E2. Q(-3, Q(-3, 0) 0) and R(6, R(6, 4)
4x – 6y+9 =0
E3. Q(2, -3) and R(-4, 5)
18x + 8y = 43
E4. Q(6, Q(6, -2) -2) and R(0, R(0, 2)
3x – 4y + 3 = 0
3x – 2y – 9 = 0
More challenges…….
E5. Given two points A(3, 2) and B(7, -4). Find the equation of the perpendicular bisector of AB.
E6. Given two points P(4, 10) and QB(-6, 0). Find the equation of the the perpendicular bisector of PQ.
3y =2x - 13
6. Coordinate Geometry
x + y = 4
10
TASK : To find the equation equation of the locus of the moving moving point P such that its distances from the points A and B are in the ratio m : n (Note : Sketch a diagram diagram to help you using the distance distance formula formula correctly) Eg 1. Let
A(A(-2,3) 2,3),, B(4, B(4,8) 8) and and P = (x, y)
m : n = 1: 2
LP PM
=
B(4, 8)
1 2
2 LK = KM
2 ( x (2)) ( y 3) = 2
2 ( x 2) 2
2
2
( y 3)
2
A(-2, 3)
( x 4) ( y 8)
= x 4) 2
2
2
( y 8
2
2
2
1
● P(x, y)
( y 3 ) = x 4 ( y 8) 2 4 x 2 16 x 16 4 y 2 24 y 36 x 2 8 x 16 y 2 16 y 64 3 x 2 3 y 2 24 x 8 y 28 0 is the equation of locus of P. 4( x 2)
E1. E1.
2
2
2
A(1, A(1, 5), B(4, (4, 2) and and m : n = 2 : 1
E2. E2.
x2+y2 – 10x – 2y + 19 = 0
E3. E3. A(1, A(1, 3), 3), B(B(-2, 2, 6) 6) and and m : n = 1 : 2
x2+y2 – 10x – 6y + 13 = 0
E4. E4.
x2+ y2 – 3x 3x – 3y 3y = 0
E5. E5. P(-1 P(-1,, 3), 3), Q( 4, -2) -2) and and m : n = 2 : 3
2
A(5, A(5, -2), 2), B(B(-4, 1) and and m : n = 1 : 2
x2+ y2 – 16x + 6y + 33 = 0
E6. E6.
2
A(1, A(1, 5), B(-4, (-4, -5) and and m : n = 3 : 2
2
x +y + 10x – 14y + 2 = 0
6. Coordinate Geometry
A(A(-3, 2), B(3, B(3, 2) and and m : n = 2 : 1
2
x +y + 16x +26y + 53 = 0
11
SPM FORMAT QUESTIONS 1. (2003) (2003) The equations equations of of two straight straight lines lines are
y
5
x
3
2. The equations equations of two straight straight lines lines are
x
1 and 5y = 3x + 24.
3
Determine whether the lines are perpendicular to each other.
y
2
4 and 3y = 2x + 6. Determine Determine
whether the lines are perpendicular to each other.
[Y]
[N]
3.(2004) 3.(2004) Diagram Diagram 4 shows a straight straight line PQ with
4. Diag Diagra ram m 5 show showss a stra straig ight ht line line RS with with the the
the equation equation
x
2
y
3
1 . Find the equation of the equation
stra straig ight ht line line perp perpen endic dicul ular ar to PQ and and pass passin ing g through the point Q.
x
6
y
4
1 . Find Find the the equa equati tion on of the the
stra straig ight ht line line perp perpen endi dicu cula larr to RS and and pass passing ing through the point S.
y
y Q
R Diagram 4 P
O
Diagram 5 S
x
O
x
[ y 2 x 3 ]
[2y = 3x - 18]
3
5. (2005) The following information refers to the equations of two straight lines, JK and RT, which are perpendicular to each other.
6. The following information refers to the equations of two straight lines, PQ and RS, which are perpendicular to each other.
JK : y = px + k RT : y = (k – 2)x + p where p and k are constants.
PQ : px + y = k RS : y = (2k –1)x + p where p and k are constants.
Express p in terms of k.
Express p in terms of k.
p
6. Coordinate Geometry
1
p
2 k
12
1 2k 1
7. (2006) Diagram 5 shows the straight line AB which is perpendicular to the straight line CB at the point B. y
8.Diagram 6 shows the straight line PQ which is perpendicul perpendicular ar to the straight straight line RQ at the point Q. y P(0, 6)
A(0, 4)
●
●
Diagram 5
B
●
x
O
x
O
Diagram 6
●
●
R
●
C
The equat equation ion of of CB is y = 2x – 1 . Find the coordinates of B.
The equat equation ion of of QR is x – y = 4 . Find the coordinates of Q
Q(5, 1)
(2, 3)
9.(2004) 9.(2004) The point A is (-1, 2) and B is (4, 6). The point P moves moves such that that PA : PB = 2 : 3. Find the the equation of locus of P. [3 marks]
2
10. The point R is (3, -5) and S is (0, 1). The The point P moves moves such such that that PR : PS = 2 : 1. Find Find the equation of locus of P. [3 marks]
2
2
[5x +5y +50x+12y+163=0]
11.The point A is (8, -2) and B is (4, 6). Find the equation of the perpendicular bisector of AB. [3 marks]
12. The point point R is (2, -3) and S is (4, (4, 5). The point point P moves such that it is always the same distance from from R and from from S. Find Find the equatio equation n of locus of P. [3 marks]
2y = x – 2
6. Coordinate Geometry
2
[x +y +2x – 6 y – 10 = 0]
x+4y = 7
13
