It is given that

y = m_{1}x + c_{1} ….. (1)

y = m_{2}x + c_{2} ….. (2)

y = m_{3}x + c_{3} ….. (3)

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

0 = (m_{2} – m_{1}) x + (c_{2} – c_{1})

(m_{1} – m_{2}) x = c_{2} – c_{1}

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 agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Find the angles between pairs of straight lines 3x – y + 5 = 0 and x – 3y + 1 = 0

Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0