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.

Asked by Abhisek | 1 year ago |  63

##### Solution :-

It is given that

y = m1x + c1 ….. (1)

y = m2x + c2 ….. (2)

y = m3x + c3 ….. (3)

By subtracting equation (1) from (2) we get

0 = (m2 – m1) x + (c2 – c1)

(m1 – m2) x = c2 – c1

So we get

$$x= \dfrac{c_2-c_1}{m_1-m_2}$$

Putting in eqn (i), we get,

$$y=\dfrac {m_1(c_2-c_1)}{m_1-m_2}+c_1$$

$$y= \dfrac{m_1c_2-m_2c_1}{m_1-m_2}$$

Putting it the eqn(iii), we get

$$y= \dfrac{m_1c_2-m_2c_1}{m_1-m_2}$$

$$y= \dfrac{m_3(c_2-c_1)+m_1c_3-m_2c_3}{m_1-m_2}$$

Rearranging, we get

$$m_1​(c_2​−c_3​)+m_2​(c_3​−c_1​)+m_3​(c_1​−c_2​)=0$$

Answered by Pragya Singh | 1 year ago

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