1ur second eqn must be 12x - 5 y -2 =0
then the answer i got is (b )
99x - 27y - 2 =0
Equation of angle bisector of the lines 3x+4y-1 =0 and 12x+5y-3=0 containing the point (1,2) is
A) 3x + 11y - 4 =0 B) 99x - 27y - 2 =0
C) 3x + 11y + 4 = 0 D) 99x + 27y - 2 = 0
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27 Answers
1
4THANKS A LOT NISHANT SIR
Sorry every one for the wrong quetion
It is from a test paper there must have been a printing mistake
But ya as Sir said this qs gave me a clear idea of how to deal with such qs
[159]
62What is the sign of the distance of the point (1,2) with respect to the lines -3x+4y-1=0? it is -3+8-1=4 which is +ve
The sign of the distance of the point (1,2) wrt 12x+5y-3 is +ve as well..
Any shift in the sector which divides the two lines will lead to one sign change.. So any point on the angular bisector in the same region as (1,2) should have either both +ve signs or both -ve
(The reason for the above comes from the graph of the two lines!)
take a point on 3x+11y-4 as (4/3,0)
The sign of the distance of the point (4/3,0) with respect to the lines -3x+4y-1=0? it is -4-1=-5 which is -ve
The sign of the distance of the point (4/3,0) wrt 12x+5y-3 is +ve
So this means that we have changed the sector only once..
So the required angular bisector will be 99x-27y-2=0
PS: All the; above calculation is based on the changed question!
62Equation of angle bisector of the lines 3x+4y-1 =0 and 12x+5y-3=0 containing the point (1,2) is
A) 3x + 11y - 4 =0 B) 99x - 27y - 2 =0
C) 3x + 11y + 4 = 0 D) 99x + 27y - 2 = 0
This is a pretty standard question, where you want to find the angular bisector..
I guess the right quesiton is "-3x+4y-1 =0 " and not "3x+4y-1 =0"
Assuming that the question is what I have written, (The important thing is to understand how to solve it rather than reaching the answer which ofcourse will fetch us marks [6])
Let us first try to see both the angular bisectors...
They are \left\{\frac{-3x+4y-1}{5}\right\}=\pm\left\{\frac{12x+5y-3}{13}\right\}
Which is same as
\\-39x+52y-13=\pm\left\{60x+25y-15 \right\} \\21x+77y-28=0 \text{ which is same as } 3x+11y-4 \\or \\99x-27y-2=0
This was the easy part.
Now how do you figure out, which of these equations stand for the one which lies in the region containing the point 1,2...
There is a simple way out here too...
take any point on the angular bisector.... (If the signs of distance are the same as that for the point 1,2 then you are done [1])
1either options given by uttara r rong or the lines...thats for sure
it will be either acute angle bisector or obtuse..
none in options is acute or obtuse
1The problem is that none of the options are either acute or obtuse angle bisectors...............then how to do it ???
1Oh me gosh !!!
Che think of such a question (which you've previously heard) :
Which of the angle bisectors contain the origin - type of thing.
It must be either acute angle bisector or obtuse angle bisector.
1well that doesnt make a sense....how can a single line contain a point ....two lines can contain a point....not one....and further der r jus 2 angle bisectors of any two intersecting lines...not more...even those r not matching with the options!
btw even if u do that wat i can think of is that the obtuse angle bisector of the given lines will contain (1,2)
1@uttara can u jus exaplin wat do u mean wen u say line contains point (1,2)
u mean to say it lies on the given line ?
if thats it then check eq of lines once again
point (1,2) lies on none of the two angle bisectors of the given lines
even u jus substitute (1,2) in the given options
none is satisfying !
1hmmm now got it . it's like "containing an origin" type of thing.
4@Pranav Prince
The equation is as given in the question
So qs is all correct
& Pritish ya thats what i was trying to tell in # 11
Anyways thanks for ure explanation
But what abt the solution ?????????
1ur second eqn must be 12x - 5 y -2 =0
then the answer i got is (b )
99x - 27y - 2 =0
39I guess(common sense, so don't flame me) that for each pair of straight lines, there will be two equations of angle bisectors, one for the smaller angle and one for the larger angle. "Containing the point (1,2)" would mean in the region of whichever angle bisector satisfies or gives < 0 on putting the point in the eqn?
4No guys The qs is correct
The qs " containing the point (1,2) " means the region containing this point
1tapas....your eqn. is right.
Uttara please verify this question again
1@TAPAS isnt da point (1,2) lying on the bisector f two equation??
139x + 52y - 13 = + or - (60x + 25y -15)
In one of the above equation 1,2 should staisfy
1m getting
33y+16x-115=0 or 33y-16x-83=0 (none of the above equation)
