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find the equation of the sides of a triangle having (4,-1) as a vertex,, if the lines x-1=0 and x-y-1=0 are the equations of two internal bisectors of its angles.

find the equation of the sides of a triangle  having (4,-1) as a vertex,, if the lines  x-1=0 and x-y-1=0 are the equations of two internal bisectors of its angles.



2 Answers

Aman Bansal
592 Points
9 years ago

Dear Pooja,

the other 2 vertices are (1,h) and (k,k+1).....Now use the fact that those lines bisect the angle between them to figure out the 2 variables h & k.

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Aman Bansal

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kunal gupta
19 Points
3 years ago
Answer :Explanation / approach:2 important points , the knowledge of which will greatly reduce the effort required to solve the problem.If 2 lines are symmetric about any x=a,x=a, then their slopes will be negative of each other. Suppose, one equation is known to us as y=mx+cy=mx+c, then the other equation will be y=−mx+dy=−mx+dIf y=x+ky=x+k is the symmetric axis, then the product of slopes should be 11. Example, the pair of lines y=4x+cy=4x+c and y=0.25x+dy=0.25x+d, will be symmetric about some y=x+ky=x+k. In other words, the coefficients of x and y get exchanged.Once, the above points are clear, all that is left is to start with some line, say AB and use the constraints like passing through A, concurrency of AC, CB and y=x−1y=x−1, and concurrency of AB, BC and x = 1.AB : y=mx+cy=mx+cPassing through A (putting x = 4 and y = -1 ), so y=mx−1−4my=mx−1−4mB : intersection of x=1x=1 and AB; so B comes out to be (1,−1−4m)(1,−1−4m)BC is symmetric with AB about x = 1, so BC : y=−mx+dy=−mx+d, but it passes through the point B(1,−1–4m)(1,−1–4m), so using this, we get y=−mx−1−2my=−mx−1−2m.Now, using the second point discussed earlier, we can write the equation for AC as follows:BC is : y=−mx−1−2my=−mx−1−2m,or y+mx+1+2m=0y+mx+1+2m=0exchanging coefficients to get AC : my+x+e=0my+x+e=0Again, this line passes through point A(4,−1)(4,−1), so using that, AC : my+x+m−4=0my+x+m−4=0Now, we use equations for BC, AC and y=x−1y=x−1, to arrive at a value of mmPutting y=x−1y=x−1, in equations BC and AC, we getx+mx+2m=0x+mx+2m=0, andmx+x−4=0mx+x−4=0xx value from the above 2 equations should be same as they correspond to a common point, (concurrence). Equating that we get :−2m1+m=41+m−2m1+m=41+mor m=−2m=−2All the triangle’s sides’ equations can now be conveniently obtained asAB : y=−2x+7y=−2x+7BC : y=2x+3y=2x+3CA : y=0.5x−3y=0.5x−3This question actually took me quite some time to figure out. Eventually it struck me that it has got to do with reflections of line about some symmetric axes. Once, I realized the importance of that in the question, the workout was simplified. I did google the part about exchanging coefficients of x and y, though

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