The equation of given lines is,
y= m 1 x+ c 1 (1)
y= m 2 x+ c 2 (2)
y= m 3 x+ c 3 (3)
Subtract equation (1) from equation (2).
0=( m 2 − m 1 )x+( c 2 − c 1 ) ( m 2 − m 1 )x=−( c 2 − c 1 ) x= −( c 2 − c 1 ) ( m 2 − m 1 ) x= ( c 2 − c 1 ) ( m 1 − m 2 )
Substitute the value of x in equation (1).
y= m 1 × ( c 2 − c 1 ) ( m 1 − m 2 ) + c 1 = m 1 c 2 − m 1 c 1 m 1 − m 2 + c 1 = m 1 c 2 − m 1 c 1 + c 1 ×( m 1 − m 2 ) m 1 − m 2 = m 1 c 2 − m 1 c 1 + c 1 m 1 − c 1 m 2 m 1 − m 2
Further solve the above expression.
y= m 1 c 2 − m 2 c 1 m 1 − m 2
The coordinate of the point of intersection of line (1) and line (2) is ( c 2 − c 1 m 1 − m 2 , m 1 c 2 − m 2 c 1 m 1 − m 2 ).
It is given in the question that line (1), line (2) and line (3) are concurrent.
Thus, the point ( c 2 − c 1 m 1 − m 2 , m 1 c 2 − m 2 c 1 m 1 − m 2 ) will also satisfy the equation of line (3).
Substitute the value of point in equation of line (3).
m 1 c 2 − m 2 c 1 m 1 − m 2 = m 3 c 2 − c 1 m 1 − m 2 + c 3 m 1 c 2 − m 2 c 1 m 1 − m 2 = m 3 c 2 − m 3 c 1 + c 3 ( m 1 − m 2 ) m 1 − m 2 m 1 c 2 − m 2 c 1 = m 3 c 2 − m 3 c 1 + c 3 m 1 − c 3 m 2 m 1 c 2 − m 2 c 1 − m 3 c 2 + m 3 c 1 − c 3 m 1 + c 3 m 2 =0
Further simplify the above expression.
m 1 ( c 2 − c 3 )+ m 2 ( c 3 − c 1 )+ m 3 ( c 1 − c 2 )=0
Hence, the condition for the three lines y= m 1 x+ c 1 , y= m 2 x+ c 2 , and y= m 3 x+ c 3 to be concurrent is m 1 ( c 2 − c 3 )+ m 2 ( c 3 − c 1 )+ m 3 ( c 1 − c 2 )=0.