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Given three straight lines 2x+11 y-5=0,2...

Given three straight lines `2x+11 y-5=0,24 x+7y-20=0,` and `4x-3y-2=0` . Then, they form a triangle one line bisects the angle between the other two two of them are parallel

A

from a triangle

B

are only concurrent

C

are concurrent with one line bisecting the angle between the other two

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
C

For the two lines
`24x+7y-20=0`
and `4x-3y-2=0,`
the angle bisectors are given by
`(24x+7y-20)/(25)=pm(4x-3y-2)/(5)`
Taking positive sign, we get
`2x+11y-5=0`
So, the given three lines are concurrent with one line bisecting the angle between the other two.
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