Question

Question 10 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

Class X1 - Maths -Straight Lines Page 233

Answer 1

Given, equation of lines are
y = m₁x + c₁--------------(1)
y = m₂x + c₂--------------(2)
y = m₃x + c₃----------------(3)

solve equations (1) and (2)
m₁x + c₁ = m₂x + c₂
x(m₁ - m₂) = c₂ - c₁
x = (c₂ - c₁)/(m₁ - m₂) put it in equation (1)
y = (m₁c₂ - m₂c₁)/(m₁ - m₂)

now, put the values of x and y in equation (3)
(m₁c₂ - m₂c₁)/(m₁ - m₂) = m₃(c₂ - c₁)/(m₁ - m₂) +c₃
(m₁c₂ - m₂c₁)/(m₁ - m₂) = {m₃c₂ - m₃c₁ + m₁c₃ - m₂c₃}/(m₁ - m₂)
(m₁c₂ - m₂c₁) = {m₃c₂ - m₃c₁ + m₁c₃ - m₂c₃}
m₁c₂ - m₂c₁ - {m₃c₂ - m₃c₁ + m₁c₃ - m₂c₃} = 0
m₁(c₂ - c₁) + m₂(c₃ - c₁) + m(c₁ - c₂) =0
hence proved

Answer 2

Answer:

hope it helpsprevent