64. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1(c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0.

64. The given eqn of the three lines arey = m1 x + c1 ______ (1)y = m2 x + c2 ______ (2)y = m3 x + c3 ______ (3)The point of intersection of (2) and (3) is given by.y - y = (m2x + c2) - (m3 x + c3)(m2 - m3) x = c3 - c2⇒x=c3-c2m2-m3.Hence, y = m2(c3-c2)(m2-m3)+c2=m2(c3-c2)+c2(m2-m3)m2-m3.=m2c3-m3c2m2-m3.ie,(c3-c2m2-m3,m2c3-m3c2m2-m3)As the three lines are concurrent, the point of intersection of (2) and (3) lies on line (1) alsoi e, m2c3-m3c2m2-m3=m1(c3-c2m2-m3)+c1⇒-m1(c2-c3)+c1(m2-m3)m2-m3=-m2c3-m3c2m2-m3.m1 (c2 - c3) - c1 (m2 - m3) + m2 c3 - m3 c2 = 0m1 (c2 - c3) - m2 c1 + m3 c1 + m2 c3 - m3 c2 = 0m1 (c2 - c3) + m2 (c3 - c1) + m3 (c1 - c2) = 0

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Ncert Solutions Maths class 11th Maths Class 11th Straight Lines

**64. If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.**

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alok kumar singh | Contributor-Level 10

2 months ago

64. The given eqⁿ of the three lines are

y = m₁ x + c₁ ______ (1)

y = m₂ x + c₂ ______ (2)

y = m₃ x + c₃ ______ (3)

The point of intersection of (2) and (3) is given by.

y - y = (m₂x + c₂) - (m₃ x + c₃)

(m₂ - m₃) x = c₃ - c₂

$\Rightarrow x = \frac{c_{3} - c_{2}}{m_{2} - m_{3}} .$

Hence, y = $\frac{m_{2} (c_{3} - c_{2})}{(m_{2} - m_{3})} + c_{2}$

$= \frac{m_{2} (c_{3} - c_{2}) + c_{2} (m_{2} - m_{3})}{m_{2} - m_{3} .}$

$= \frac{m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}} .$

$i e, (\frac{c_{3} - c_{2}}{m_{2} - m_{3}}, \frac{m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}})$

As the three lines are concurrent, the point of intersection of (2) and (3) lies on line (1) also

i e, $\frac{m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}} = m_{1} (\frac{c_{3} - c_{2}}{m_{2} - m_{3}}) + c_{1}$

$\Rightarrow \frac{- m_{1} (c_{2} - c_{3}) + c_{1} (m_{2} - m_{3})}{m_{2} - m_{3}} = \frac{- m_{2} c_{3} - m_{3} c_{2}}{m_{2} - m_{3}} .$

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

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

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

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**64. If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.**

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