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Question

Prove that the three straight lines whose equations are 15x18y+1=0,12x+10y3=0, and 6x+66y11=0 all meet in a point.
Show also that the third line bisects the angle between the other two.

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Solution

15x18y+1=0.......(i)12x+10y3=0..........(ii)12x=310yx=310y12

Substituting x in (i), we get

15(310y12)18y+1=01550y72y+44=019122y=0y=19122x=310y12=310(19122)12x=366190(122)12=22183

So the point of intersection is P(22183,19122)

6x+66y11=0......(iii)

Substituting P in (iii)

6×22183+66×1912211=04461+6276111=06716111=01111=00=0

P also satisfies the equation of third line, so they all meet in a point .

Equation of angle bisector between 15x18y+1=0 and 12x+10y3=0 is

15x18y+1(15)2+(18)2=±12x+10y3(12)2+(10)215x18y+1549=±12x+10y324415x18y+1361=±12x+10y32612(15x18y+1)=±3(12x+10y3)30x36y+2=±(36x+30y3)30x36y+2=36x+30y96x+66y11=0

which is same as (ii)

Hence, third line bisects the angle between the other two.


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