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Question

If three lines whose equations are y=m1x+c1,y=m2x+c2 and y=m3+c3 are concurrent, then show that m1(c2c3)+m2(c3c1)+m3(c1c2)=0

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Solution

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=(m2m1)x+(c2c1)
(m1m2)x=c2c1
x=c2c1m1m2
On substituting this value of x in (1), we obtain
y=m1{c2c1m1m2}+c1
y=m1c2m1c1m1m2+c1
y=m1c2m1c1+m1c1m2c1m1m2
y=m1c2m2c1m1m2
{c2c1m1m2,m1c2m2c1m1m2} 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).
m1c2m2c1m1m2=m3{c2c1m1m2}+c3
m1c2m2c1m1m2=m3c2m3c1+c3m1c3m2m1m2
m1(c2c3)+m2(c3c1)+m3(c1c2)=0
Hence, m1(c2c3)+m2(c3c1)+m3(c1c2)=0

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