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✴✴✴class 11✴✴✴✴
Answers
hey ya....
https://www.cbsetuts.com/ncert-solutions-class-11-computer-science-python-conditional-looping-constructs/
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