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प्रश्न
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.
उत्तर
The equations of the given lines are
y = m1x + c1 …(1)
y = m2x + c2 …(2)
y = m3x + c3 …(3)
On subtracting equation (1) from (2), we obtain
0 = (m2 − m1)x + (c2 − c1)
= (m1 − m2)x = c2 − c1
= `x = (c_2 - c_1)/(m_1 - m_2)`
On substituting this value of x in (1), we obtain
`y = m_1 ((c_2 - c_1)/(m_1 - m_2)) + c_1`
`y = ((m_1c_2 - m_1c_1)/(m_1 - m_2)) + c_1`
`y = (m_1c_2 - m_1c_1 + m_1c_1 - m_2c_1)/(m_1 - m_2)`
`y = (m_1c_2 - m_2c_1)/(m_1 - m_2)`
∴ `((c_2 - c_1)/(m_1 - m_2), (m_1c_2 - m_2c_1)/(m_1 - m_2))` is the point of intersection of lines (1) and (2).
It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).
= `(m_1c_2 - m_2c_1)/(m_1 - m_2) = m_3 ((c_2 - c_1)/(m_1 - m_1)) + c_3`
= `(m_1c_2 - m_2c_1)/(m_1 - m_2) = (m_3c_2 - m_3c_1 + c_3m_1 - c_3m_2)/(m_1 - m_2)`
= m1c2 - m2c1 - m3c2 + m3c1 - c3m1 + c3m2 = 0
= m1 (c2 - c3) + m2 (c3 - c1) + m3 (c1 - c2) = 0
Hence, m1 (c2 - c3) + m2 (c3 - c1) + m3 (c1 - c2) = 0
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