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Exercise 4
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Calculus and Analytic Geometry, MATHEMATICS 12

Merging man and maths

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6

Point of intersection of lines
Let l1 : a1 x + b1 y + c1 = 0
l2 : a2 x + b2 y + c2 = 0 be non-parallel lines
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Then
a1 x1 + b1 y1 + c1 = 0
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(ii )
Solving (i ) and (ii ) simultaneously, we have
x1
− y1
1
=
=
b1c2 − b2c1 a1c2 − a2c1 a1b2 − a2b1
x1
1
− y1
1
⇒
=
and
=
b1c2 − b2c1 a1b2 − a2b1
a1c2 − a2c1 a1b2 − a2b1
bc −b c
a c − a2c1
⇒ x1 = 1 2 2 1 and y1 = − 1 2
a1b2 − a2b1
a1b2 − a2b1
 bc −b c
a c − a2c1 
Hence  1 2 2 1 , − 1 2
 is the point of intersection of l1 and l2
...

Let l1 : a1 x + b1 y + c1 = 0
l2 : a2 x + b2 y + c2 = 0
Then equation of line passing through the point of intersection of l1 and l2 is
l1 + k l2 = 0 , where k is constant
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e
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Now if ( x, y ) is the point of intersection of l1 and l2 then

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4 - 2

x
−y
1
=
=
(−2)(2) − (−1)(1) (1)(2) − (2)(1) (1)(−1) − (2)(−2)
x
−y
1
⇒
=
=
−4 + 1 2 − 2 −1 + 4
x −y 1
⇒
=
=
−3 0 3
x 1
−y 1
⇒
=
=
and
−3 3
0 3
−3
0
⇒ x=
and
y=−
3
3
⇒ x = −1
and
y=0
Hence ( −1,0 ) is the point of intersection
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Now if ( x, y ) is the point of intersection of l1 and l2 then
x
−y
1
=
=
−1 − 24 −3 − 12 6 − 1
x
−y 1
⇒
=
=
−25 −15 5
x
1
−y 1
⇒
=
and
=
−25 5
−15 5
−25
15
⇒ x=
= −5 and y = = 3
5
5
Hence ( −5,3) is the point of intersection
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Question # 2
Find an equation of the line through
(i)
the point ( 2, −9 ) and the intersection of the lines 2 x + 5 y − 8 = 0 and
3x − 4 y − 6 = 0
(ii) the intersection of the lines
x− y−4=0
7 x + y + 20 = 0
6 x + y − 14 = 0
(a) Parallel
(ii) Perpendicular
to the line 6 x + y − 14 = 0
(iii) through the intersection of the lines x + 2 y + 3 = 0, 3x + 4 y + 7 = 0
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And making equal intercepts on the axes
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(i)
Since ( 2, −9 ) lies on (i) therefore put x = 2 and y = −9 in (i)
2(2) + 5(−9) − 8 + k ( 3(2) − 4(−9) − 6 ) = 0
⇒ 4 − 45 − 8 + k ( 6 + 36 − 6 ) = 0
⇒ − 49 + 36k = 0
49
⇒ 36k = 49
⇒ k=
36
Putting value of k in (i)
49
2 x + 5 y − 8 + ( 3x − 4 y − 6 ) = 0
36
⇒ 72 x + 180 y − 288 + 49 ( 3 x − 4 y − 6 ) = 0
× ing by 36
⇒ 72 x + 180 y − 288 + 147 x − 196 y − 294 = 0
⇒ 219 x − 16 y − 582 = 0 is the required equation
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(i)
⇒ (1 + 7 k ) x + ( −1 + k ) y + ( − 4 + 20k ) = 0
1 + 7k
Slope of l4 = m1 = −
−1 + k
6
Slope of l3 = m2 = − = −6
1
(a) If l3 and l4 are parallel then
m1 = m2
1 + 7k
⇒ −
= −6
−1 + k
⇒ 1 + 7 k = 6 ( −1 + k ) ⇒ 1 + 7 k = −6 + 6k
⇒ 7 k − 6k = −6 − 1
⇒ k = −7
Putting value of k in (i)
x − y − 4 − 7 ( 7 x + y + 20 ) = 0
⇒ x − y − 4 − 49 x − 7 y − 140 = 0
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(b)

⇒ − 48 x − 8 y − 144 = 0
⇒ 6 x + y + 18 = 0
is the required equation
If l3 and l4 are ⊥ then
m1m2 = −1
 1 + 7k 
⇒ −
 ( −6 ) = −1
k
−
1
+


⇒ 6 (1 + 7 k ) = − ( −1 + k )
⇒ 6 + 42k = 1 − k
⇒ 42k + k = 1 − 6

⇒ 43k = −5

⇒ k =−

5
43

Putting in (i) we have
5
( 7 x + y + 20 ) = 0
43
43 x − 43 y − 172 − 5 ( 7 x + y + 20 ) = 0
43 x − 43 y − 172 − 35 x − 5 y − 100 = 0
8 x − 48 y − 272 = 0
x − 6 y − 34 = 0 is the required equation
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Question # 3
Find an equation of the line through the intersection of
16 x − 10 y − 33 = 0 ; 12 x + 14 y + 29 = 0 and the intersection of
x − y + 4 = 0 ; x − 7y + 2 = 0
Solution
Let l1 : 16 x − 10 y − 33 = 0
l2 : 12 x + 14 y + 29 = 0
l3 : x − y + 4 = 0
l4 : x − 7 y + 2 = 0
For point of intersection of l1 and l2
x
−y
1
=
=
−290 + 462 464 + 396 224 + 120
x
−y
1
⇒
=
=
172 860 334
x
1
−y
1
⇒
=
and
=
172 334
860 334
172 1
860
5
⇒ x=
=
and
y=−
=−
334 2
334
2
1 5
⇒  , −  is a point of intersection of l1 and l2
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x
−y
1
=
=
−2 + 28 2 − 4 −7 + 1
x −y 1
⇒
=
=
26 −2 −6
x
1
−y 1
⇒
=
and
=
26 −6
−2 −6
26
13
2
1
⇒ x=
=−
and
y=
=−
−6
3
−6
3
 13 1 
⇒  − , −  is a point of intersection of l3 and l4
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Three Concurrent Lines
Suppose l1 : a1 x + b1 y + c1 = 0
l2 : a2 x + b2 y + c2 = 0
l3 : a3 x + b3 y + c3 = 0
If l1 , l2 and l3 are concurrent (intersect at one point) then
a1 b1 c1
a2 b2 c2 = 0
a3 b3 c3
See proof on book at page 208

l1

l2
l3

Question # 4
Find the condition that the lines y = m1 x + c1 ; y = m2 x + c2 and y = m3 x + c3 are
concurrent
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Question # 5
Determine the value of P such that the lines 2 x − 3 y − 1 = 0 , 3 x − y − 5 = 0
and 3 x + py + 8 = 0 meet at a point
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e
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If α1 and α 2
be inclinations and m1 and m2 be slopes of
lines l1 and l2 respectively, Let θ be a angle
from line l1 to l2 then θ is given by
m − m1
tan θ = 2
1 + m1m2

Y

l2

l1
ψ
θ

α1

α2
X

See proof on book at page 219

Question # 6
Show that the lines 4 x − 3 y − 8 = 0, 3 x − 4 y − 6 = 0 and x − y − 2 = 0 are
concurrent and the third-line bisects the angle formed by the first two lines
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4 4
l2
Slope of l1 = m1 = − =
−3 3
3 3
θ2
Slope of l2 = m2 = −
=
−4 4
θ1
1
Slope of l3 = m3 = − = 1
−1
Now let θ1 be angle from l1 to l3 and θ 2 be a
angle from l3 to l2
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l3

l1

( )

( )

Question # 7
The vertices of a triangle are A ( −2,3) , B ( −4,1) and C ( 3,5 )
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(i) Centroid of triangle is the intersection of medians and is given by
 x1 + x2 + x3 y1 + y2 + y3 
,


3
3


 −2 − 4 + 3 3 + 1 + 5 
 −3 9 
=
,
 =  ,  = ( −1,3)
3
3 

 3 3
Hence ( −1,3) is the centroid of the triangle
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1− 3
−2
Slope of AB = m1 =
=
=1
−4 + 2 −2
5 −1 4
Slope of BC = m2 =
=
3+ 4 7
Since altitudes are ⊥ to sides therefore
1
1
Slope of altitude on AB = − = − = −1
m1
1

A

B

C

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1
1
7
=−
=−
4
4
m2
7
Equation of altitude on AB with slope −1 from C ( 3,5 )
y − 5 = −1( x − 3)
⇒ y − 5 = −x + 3 ⇒ x − 3 + y − 5 = 0
⇒ x + y − 8 = 0 ………
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(ii)
For point of intersection of (i) and (ii)
x
−y
1
=
=
2 + 32 2 + 56 4 − 7
x −y 1
⇒
=
=
34 58 −3
x
1
−y 1
⇒
=
and
=
34 −3
58 −3
34
58 58
⇒ x=−
and y = − =
3
−3 3
 34 58 
Hence  − ,  is orthocentre of triangle ABC
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Let D and E are midpoints of side AB and BC respectively
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(iii)
7
 1 
Now equation of ⊥ bisector having slope − through E  − ,3 
4
 2 
7
1
7
y − 3 = −  x +  ⇒ 4 y − 12 = −7 x −
4
2
2
7
17
⇒ 7 x + + 4 y − 12 = 0 ⇒ 7 x + 4 y − = 0
2
2
⇒ 14 x + 8 y − 17 = 0 ……………
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6
 6
 34 58 
 25 31 
Now to check ( −1,3) ,  − ,  and  , −  are collinear, let
6
 3 3
 6
−1
− 34

3
1
58
1
3
3
25
− 31 1
6
6
 58 31   34 25   1054 1450 
= −1 +  − 3  − −  + 1
−

6  3
6   18
18 
 3
 49   31 
= −1  − 3  −  + 1( −22 )
 2   2
49 93
= − + − 22 = 0
2
2
Hence centroid, orthocentre and circumcentre of triangle are collinear
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If so, find the point where they meet
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For point of concurrency, we find intersection of l1 and l2 (You may choose any
two lines)
x
−y
1
=
=
18 − 32 −24 + 24 −16 + 9
x
−y 1
⇒
=
=
−14 0 −7
x
1
−y 1
=
and
=
⇒
−14 −7
0 −7
−14
0
⇒ x=
= 2 and y = −
=0
−7
−7
Hence ( 2,0 ) is the point of concurrency
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Also find measures of the angles of
the triangle
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4
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5 
 5
For point of intersection of l2 and l3
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5 5 
Now for point of intersection of l1 and l3
x
−y
1
=
=
8 + 6 −4 + 12 1 + 4
x −y 1
⇒
=
=
14 8 5
x 1
−y 1
⇒
=
and
=
14 5
8 5
14
8
⇒ x=
and
y=−
5
5
 14 8 
⇒  , −  is the point of intersection of l1 and l3
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We say these vertices as A, B and C respectively
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Then

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5

3− 1
m1 − m3
2
tan α =
=
1 + m1m3 1 + (3) 1

( 2)

⇒ α = tan −1 (1)

Now

tan β =

2 =1
5
2

⇒ α = 45

m2 − m1
−2 − 3
−5
=
=
=1
1 + m2 m1 1 + (−3) ( 3) −5

⇒ β = tan −1 (1) ⇒

Now

=

β = 45

5
1 +2
m3 − m2
2
=
= 2 =∞
tan γ =
1 + m3m2 1 + 1 ( −2 ) 0
2

( )

⇒ γ = tan −1 (∞)

⇒ γ = 90
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4
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4
6
⇒ θ = 180 − 49
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6

Now acute angle between lines = 180 − 130
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4
(b)
(c)

Do yourself as above
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69

(d)

( )
( )( )

Also acute angle between lines = 78
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4
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Then
−3 − 11 −14 7
y
m1 =
=
=
−6 + 2
−4 2
A
−9 + 3 −6
3
m2 =
=
=−
4 + 6 10
5
α
m3
m
1
11 + 9 20
10
m3 =
=
=−
−2 − 4 −6
3
x
β
γ
Let α , β and γ denotes angles of triangle at vertex
m2
B
C
A, B and C respectively
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64
 64 
7 − −3
m1 − m2
2
5
tan β =
=
1 + m1m2 1 + 7
−3
2
5
7 +3
41
2
5
10 = − 41 × 10 = − 41
=
=
11
10 11
1 − 21
−11
10
10
41
41
⇒ − tan β =
⇒ tan (180 − β ) =
∵ tan(180 − θ ) = − tan θ
11
11
 41 
⇒ 180 − β = tan −1   = 74
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98
⇒ β = 105
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34
 45 
(b) Given vertices A ( 6,1) , B ( 2,7 ) and C ( −6, −7 )
Let m1 , m2 and m3 denotes the slopes of side AB , BC and CA respectively
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Then
C
2 − −3
m3 − m1
3
2
tan α =
=
1 + m3m1 1 + 2
−3
3
2
13
2 +3
3
2
=
= 6 =∞
1−1
0

( )
( )( )

⇒ α = tan −1 ( ∞ )

⇒ α = 90

−3 −7
m1 − m2
2
4
tan β =
=
3
7
1 + m1m2 1 + −
2
4
−13
−13
4
4 = 13 × 8 = 2
=
=
4 13
1 − 21
−13
8
8
⇒ β = tan −1 (2) ⇒ β = 63
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57
2

( )( )

(c)

Do yourself as above
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Question # 12
Find the interior angles of the quadrilateral whose vertices are A ( 5, 2 ) , B ( −2,3) ,
C ( −3, −4 ) and D ( 4, −5 )
Solution
Given vertices are A ( 5, 2 ) , B ( −2,3) , C ( −3, −4 ) and D ( 4, −5 )
Let m1 , m2 , m3 and m4 be slopes of side AB, BC , CD and DA
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Then
50
7− −1
7+ 1
m4 − m1
7
7
tan α =
=
=
= 7 =∞
1
1−1
1 + m4 m1 1 + ( 7 ) −
0
7

(

)

(

⇒ α = tan −1 ( ∞ )

)

⇒ α = 90

−1 −7
− 50
m1 − m2
7
7 =∞
tan β =
=
=
1 + m1m2 1 + − 1 ( 7 )
0
7

(

⇒ β = tan −1 ( ∞ )

⇒

)

β = 90

(

7− −1
m2 − m3
7
tan γ =
=
1 + m2 m3 1 + ( 7 ) − 1

(

⇒ γ = tan −1 ( ∞ )

)
7

50
7
=
= 7 =∞
1−1
0
7+ 1

)

⇒ γ = 90

−1 −7
− 50
m3 − m4
7
7 =∞
tan δ =
=
=
1 + m3m4 1 + − 1 ( 7 )
0
7

(

⇒ δ = tan −1 ( ∞ )

)

⇒ δ = 90

Trapezium
If any two opposite sides of the quadrilateral are parallel
then it is called trapezium
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Solution
Given vertices are A ( −1, −1) , B ( −3,0 ) , C ( 3,7 ) and D (1,8 )
Let m1 , m2 , m3 and m4 be slopes of side AB, BD, DC and CA
...

Now suppose α , β , γ and δ are angels of
quadrilateral at vertices A, B, C and D respectively
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Now for point of intersection of l2 and l3
x
−y
1
=
=
3 + 82 30 + 123 20 − 3
x −y 1
⇒
=
=
85 153 17
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x

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x −y 1
=
=
85 153 17
x
1
−y 1
⇒
=
and
=
85 17
153 17
85
153
⇒ x=
= 5 and
y=−
= −9
17
17
⇒ ( 5, −9 ) is the point of intersection of l2 and l3
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Now area of triangle having vertices ( 3,11) , ( 5, −9 ) and (1, −3) is given by:
⇒

3 11 1
1
5 −9 1
2 1 −3 1
1
= 3 ( −9 + 3) − 11( 5 − 1) + 1( −15 + 9 )
2
1
1
= 3 ( −6 ) − 11( 4 ) + 1( −6 ) = −18 − 44 − 6
2
2
1
1
= −68 = (68) = 34 sq
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Find the centre of the
circum centre of the triangle?
Solution
Same Question # 7(c)
Question # 16
Express the given system of equations in matrix form
...

(a) x + 3 y − 2 = 0 ;
2 x − y + 4 = 0 ; x − 11 y + 14 = 0
(b) 2 x + 3 y + 4 = 0 ;
x − 2 y − 3 = 0 ; 3x + y − 8 = 0
(c) 3 x − 4 y − 2 = 0 ;
x + 2 y − 4 = 0 ; 3x − 2 y + 5 = 0

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Solution
(a)

x + 3y − 2 = 0
2x − y + 4 = 0
x − 11 y + 14 = 0

In matrix form
 1 3 −2   x   0 
 2 −1 4   y  =  0 
1 −11 14   1  0 

   
Coefficient matrix of the system is
 1 3 −2 
A =  2 −1 4 
1 −11 14 


⇒ A = 1(−14 + 44) − 3(28 − 4) − 2(−22 + 1)
= 1(30) − 3(24) − 2(−21)
= 30 − 72 + 42 = 0
Hence given lines are concurrent
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(c)

Do yourself as above

Question # 17
Find a system of linear equations corresponding to the given matrix form
...

1 0 −1  x  0 
(a)  2 0 1   y  = 0
 0 −1 2   1   0 

   

1 1 2   x  0
(b)  2 4 −3  y  = 0 
 3 6 −5   1   0 

   

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Solution
(a)
1 0 −1  x  0 
 2 0 1   y  = 0 
 0 −1 2   1   0 

   
 x + 0 − 1  0
 x − 1  0 
⇒  2 x + 0 + 1 = 0  ⇒  2 x + 1  = 0 
 0 − y + 2  0
 − y + 2  0 

  

  
Equating the elements
x −1 = 0
2x + 1 = 0
−y + 2 = 0
are the required equation of lines
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Do yourself as above
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4 (Page 223)
Calculus and Analytic Geometry Mathematic 12
Punjab Textbook Board, Lahore
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Updated: October,10,2017
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