1@Debo-did u try it?
Sorry, the right question is:- "The sides AB & AC are bisected at right angles by the pair of straight lines y2-5xy-4x2=0. The base BC of the triangle passes through a fixed point (4,5). Here, AB, AC & BC are the sides of a triangle.
Determine the locus of its vertex A."
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13 Answers
19just gave it a dash earlier !!! i have not done it completely, though ! ;) ;)
11This is what's coming to my head in 1 shot....
Eqn of BC is (x-4)=m(y-5).........(2)
mAB=-1
Equation of AB comes out to be y+x=h+k,...(1) where (h,k) represent the vertex A.
Solving (1) and (2), the x and y co-ordinates of B turns out to be m(h+k-5)+4m+1 & h+k+5m-4m+1
Since AB is bisected by y=4x, so h+xAB2=4h+yAB2
Get m in terms of h and k from there and thus get co-ordinates of B solely in terms of (h,k)....Now observe that (0.0) is the circum-centre of ABC, so xB2+yB2=h2+k2.......That's the reqd locus friend.....no polar co-ordinates reqd.
1@soumik-- as u took the slope of AB to be = -1, so how can it be bisected by y=4x??
The pair of straight lines can be separated into y=x & y=4x.
Now, AB & AC are _|_ to these lines.
So, (Slope of AB)(1) = -1 & (Slope of BC)(4) = -1 or you can take it the other way round.....
So, according to you BC is bisected by the line y=4x not AB if you take the slope of AB = 1...........
24Bisected at rite angles..this means they are altitudes..and they intesect at otrhocentre..(i,e point of intersection os pair of lines)
form ther ewe can get locus of A..
am i on rite track ??
1they are not altitudes; they are infact _|_ bisectors of lines AB & AC......its a very simple question......no funda of Triangles is reqd here.....just simple straight lines is enough!
Now try it out......I would feel happy to see if anyone can come up with some other method of solving it.......but first try to solve it with any method..........
21wat i got is those lines are pependicular bisectors and since they intersect at origin so circumcenter of this triangle ABC must be at origin.....
1ya thats right.......
they indeed intersect at origin......so what do u infer from that??
1The Solution is as follows:-
Let the coordinates of A(h, k), B(x1, y1) & C(x2, y2).
(Slope of AB)(4)=-1.........(i)
(Slope of BC)(1)=-1.........(ii)
Mid-point of AB satisfies y=4x............(iii)
Mid-point of BC satisfies y=x..............(iv)
Find coordinates of B & C from the above 4 eqns in the terms of h & k.
Since, BC passes through (4, 5)
=>Area of triangle formed by (4, 5), B & C=0
=> the reqd eqn of locus of A.
Replace (h, k) by (x, y).