Manual 1 Gvv Python 2d

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1

JEE Problems in Linear
Algebra: 2D
G V V Sharma∗

Contents

lam 1 = np.linspace(0,1,len)

1

Line

1

2

Medians of a Triangle

2

3

Altitudes of a Triangle

3

4

Angle Bisectors of a Triangle

4

Abstract—This manual introduces matrix computations
using python and the properties of a triangle.

1 Line
1.1 Let

!
!
!
−2
1
4
A=
,B =
,C =
.
−2
3
−1

(1)

Draw ∆ABC.
Solution: The following code yields the desired plot in Fig. 1.1
#Code by GVV Sharma
#January 28, 2019
#released under GNU GPL
import numpy as np
import matplotlib.pyplot as plt
#if using termux
import subprocess
import shlex
#end if
A = np.array([−2,−2])
B = np.array([1,3])
C = np.array([4,−1])
len =10
*The author is with the Department of Electrical Engineering,
Indian Institute of Technology, Hyderabad 502285 India e-mail:
gadepall@iith.ac.in. All content in this manual is released under GNU
GPL. Free and open source.

x AB = np.zeros((2,len))
x BC = np.zeros((2,len))
x CA = np.zeros((2,len))
for i in range(len):
temp1 = A + lam 1[i]∗(B−A)
x AB[:,i]= temp1.T
temp2 = B + lam 1[i]∗(C−B)
x BC[:,i]= temp2.T
temp3 = C + lam 1[i]∗(A−C)
x CA[:,i]= temp3.T
#print(x
plt.plot(x
plt.plot(x
plt.plot(x

AB[0,:],x
AB[0,:],x
BC[0,:],x
CA[0,:],x

AB[1,:])
AB[1,:],label=’$AB$’)
BC[1,:],label=’$BC$’)
CA[1,:],label=’$CA$’)

plt.plot(A[0], A[1], ’o’)
plt.text(A[0] ∗ (1 + 0.1), A[1] ∗ (1 − 0.1) , ’
A’)
plt.plot(B[0], B[1], ’o’)
plt.text(B[0] ∗ (1 − 0.2), B[1] ∗ (1) , ’B’)
plt.plot(C[0], C[1], ’o’)
plt.text(C[0] ∗ (1 + 0.03), C[1] ∗ (1 − 0.1) ,
’C’)
plt.xlabel(’$x$’)
plt.ylabel(’$y$’)
plt.legend(loc=’best’)
plt.grid() # minor
#if using termux
plt.savefig(’../figs/triangle.pdf’)
plt.savefig(’../figs/triangle.eps’)
subprocess.run(shlex.split(”termux−open ../
figs/triangle.pdf”))
#else
#plt.show()

2

B

3

def dir vec(AB):
return np.matmul(AB,dvec)

AB
BC
CA

def norm vec(AB):
return np.matmul(omat,np.matmul(AB,dvec
))

2

y

1

A = np.array([−2,−2])
B = np.array([1,3])
dvec = np.array([−1,1])
omat = np.array([[0,1],[−1,0]])
AB =np.vstack((A,B)).T

0
C

−1
−2

A
−2

−1

0

1
x

2

3

4

print (dir vec(AB))
print (norm vec(AB))

Fig. 1.1

1.5 Find the equations of BC and CA
1.2 Find the equation of AB.
Solution: The desired equation is obtained as
AB :

x = A + λ1 (B − A)
!
!
2
3
=−
+ λ1
2
5

(2)
(3)

Alternatively, the desired equation is


5 −3 (x − A) = 0 (4)



 2!
=⇒ 5 −3 x = − 5 −3
= −4
(5)
2
1.3 Find the direction vector and the normal vector
for AB
Solution: Let
!


−2 1
T AB = A B =
(6)
−2 3
m = B − A = T AB

=⇒ n =

!

0 1
5
m=
−1 0
−3

!

(8)

def mid pt(B,C):
D = (B+C)/2
return D

(9)

1.4 Write a python code for computing the direction and normal vectors.
import numpy as np

C+A
A+B
B+C
,E =
,F =
(10)
2
2
2
The following code computes the values resulting in
!
!
!
2.5
1
−0.5
D=
,E =
,F =
, (11)
1
−1.5
0.5
D=

(7)

The normal vector n is defined as
nT m = 0

2.1 Find the coordinates of D, E and F of the
mid points of AB, BC and CA respectively for
∆ABC.
Solution: The coordinates of the mid points
are given by

#This program calculates the mid point
between
#any two coordinates
import numpy as np
import matplotlib.pyplot as plt

The direction vector of AB is
!
!
−1
3
=
1
5

2 Medians of a Triangle

A = np. matrix(’−2;−2’)
B = np. matrix(’1;3’)
C = np. matrix(’4;−1’)
print(mid pt(B,C))
print(mid pt(C,A))
print(mid pt(A,B))

3

2.2 Find the equations of AD, BE and CF. These
lines are the medians of ∆ABC
Solution: Use the code in Problem 1.4.
2.3 Find the point of intersection of AD and CF.
Solution: Let the respective equations be
nT1 x = p1 and
nT2 x = p2

(12)
(13)

This can be written as the matrix equation
!
nT1
x=p
(14)
nT2
where

=⇒ N T x = p

(15)



N = n1 n2 ,

(16)

The point of intersection is then obtained as
 −1
x = NT p
(17)
= N −T p

(18)

The following code yields the point of intersection
!
1
G=
(19)
0

D = mid pt(B,C)
F = mid pt(A,B)
AD =np.vstack((A,D)).T
CF =np.vstack((C,F)).T
dvec = np.array([−1,1])
omat = np.array([[0,1],[−1,0]])
print(line intersect(AD,CF))
2.4 Using the code in Problem 2.3, verify that G
is the point of intersection of BE, CF as well
as AD, BE. G is known as the centroid of
∆ABC.
2.5 Graphically show that the medians of ∆ABC
meet at the centroid.
2.6 Verify that
G=

A+B+C
3

(20)

3 Altitudes of a Triangle
#This program calculates the
#intersection of AD and CF
import numpy as np
def mid pt(B,C):
D = (B+C)/2
return D
def norm vec(AB):
return np.matmul(omat,np.matmul(AB,dvec
))
def line intersect(AD,CF):
n1=norm vec(AD)
n2=norm vec(CF)
N =np.vstack((n1,n2))
p = np.zeros(2)
p[0] = np.matmul(n1,AD[:,0])
p[1] = np.matmul(n2,CF[:,0])
return np.matmul(np.linalg.inv(N),p)
A = np.array([−2,−2])
B = np.array([1,3])
C = np.array([4,−1])

3.1 In ∆ABC, Let P be a point on BC such that
AP ⊥ BC. Then AP is defined to be an altitude
of ∆ABC.
3.2 Find the equation of AP.
3.3 Find the equations of the altitudes BQ and CR.
3.4 Find the point of intersection of AP and BQ.
Solution: Using the code in Problem 2.3, the
desired point of intersection is
!
1.407
H=
(21)
0.56
Interestingly, BQ and CR also intersect at the
same point. Thus, the altitudes of a triangle
meet at a single point known as the orthocentre
3.5 Find P, Q, R.
Solution: P is the intersection of AP and BC.
Thus, the code in Problem 2.3 can be used to
find P. The desired coordinates are
!
!
!
2.32
1.73
0.03
P=
,Q =
,R =
(22)
1.24
−1.38
1.38
3.6 Draw AP, BQ and CR and verify that they meet
at H.

4

4 Angle Bisectors of a Triangle
4.1 In ∆ABC, let U be a point on BC such that
∠BAU = ∠CAU. Then AU is known as the
angle bisector.
4.2 Find the length of AB, BC and CA
Solution: The length of CA is given by
CA = kC − Ak

It is easy to verify that even BV and CW meet
at the same point. I is known as the incentre
of ∆ABC.
4.6 Draw AU, BV and CW and verify that they
meet at a point I.
4.7 Verify that
I=

(23)

BC.A + CA.B + AB.C
AB + BC + CA

(31)

The following code calculates the respective
values as
AB = 5.83, BC = 5, CA = 6.08

(24)

nT (x − A) = 0

#This program calculates the distance
between
#two points
import numpy as np
import matplotlib.pyplot as plt

IX =

=

knk

IX = IY = IZ = r
r is known as the inradius of ∆ABC.
4.10 Draw the incircle of △ABC
4.11 Draw the circumcircle of △ABC

4.3 If AU, BV and CW are the angle bisectors, find
the coordinates of U, V and W.
Solution: Using the section formula,

CA
.B + A
BC
CA
+1
BC

nT (I − A)

(33)

Verify through a Python script.
4.9 If IY ⊥ BC and IZ ⊥ CA, verify that

print (np.linalg.norm(A−B))

AW.B + W B.A
=
AW + W B

(32)

show that

A = np.array([−2,−2])
B = np.array([1,3])
C = np.array([4,−1])

W=

4.8 Let the perpendicular from I to AB be IX. If
the equation of AB is

AW
.B + A
WB
AW
+1
WB

(25)
(26)

CA × B + BC × A
(27)
BC + CA
a×A+b×B
(28)
=
a+b
where a = BC, b = CA, since the angle bisector
has the property that
=

AW CA
=
(29)
W B AB
4.4 Write a program to find U, V, W.
4.5 Find the intersection of AU and BV.
Solution: Using the code in Problem 2.3, the
desired point of intersection is
!
1.15
I=
(30)
0.14

(34)



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