Computer Science, asked by NidhraNair, 1 year ago

help me with this question¡¡!!
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©lass 11 python
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Thank you;)

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Answers

Answered by Anonymous
1

Hi,

What is type()?

In python type() function is used to check the data type of given argument. it returns type of the given object.

So here is your answer

a)<class 'int'>

b)<class 'int'>

c)SyntaxError: invalid syntax

d)<class 'str'>

e)<class 'float'>

f)<class 'int'>

g)<class 'float'>

h)<class 'float'>

i)<class 'float'>

You can also check screenshot to see real output.

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NidhraNair: thank you sooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooo sooooooooo sooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooo sooooooooo sooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooo sooooooooo sooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooo sooooooooo sooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooooooooooooo sooooooooo sooooooooo much
NidhraNair: ;)
Answered by shivamtiwari26
0

Answer:

As these three lines are Concurrent : they all intersect at a single point

⇒ The Given three lines have Common point of Intersection

Let us find the Point of intersection of any two lines :

1st line : y = m₁x + c₁

substituting the value of y in the 2nd line : y = m₂x + c₂

⇒ m₁x + c₁ = m₂x + c₂

⇒ x(m₁ - m₂) = c₂ - c₁

⇒ x = \frac{c_{2} - c_{1}}{m_1 - m_2}x=

m

1

−m

2

c

2

−c

1

⇒ y = \frac{m_1(c_2 - c_1)}{m_1 - m_2} + c_1y=

m

1

−m

2

m

1

(c

2

−c

1

)

+c

1

substituting the x and y values in 3rd line : y = m₃x + c₃

⇒ \frac{m_1(c_2 - c_1)}{m_1 - m_2} + c_1 = \frac{m_3(c_2 - c_1)}{m_1 - m_2} + c_3

m

1

−m

2

m

1

(c

2

−c

1

)

+c

1

=

m

1

−m

2

m

3

(c

2

−c

1

)

+c

3

⇒ \frac{m_1(c_2 - c_1)}{m_1 - m_2} = \frac{m_3(c_2 - c_1)}{m_1 - m_2} + c_3 - c_1

m

1

−m

2

m

1

(c

2

−c

1

)

=

m

1

−m

2

m

3

(c

2

−c

1

)

+c

3

−c

1

⇒ m₁(c₂ - c₁) = m₃(c₂ - c₁) + (m₁ - m₂)(c₃ - c₁)

⇒ m₃(c₂ - c₁) + m₁(c₃ - c₁ - c₂ + c₁) - m₂(c₃ - c₁) = 0

⇒ m₃(c₂ - c₁) + m₁(c₃ - c₂) - m₂(c₃ - c₁) = 0

⇒ m₃(c₁ - c₂) + m₁(c₂ - c₃) + m₂(c₃ - c₁) = 0

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