1jesh frm my solved equation the centre of the circle is (0,c) yaar.
1) The line Ax+By+C=0 cuts the x-axis at P and y - axis at Q.Find the co-ordinates of the ortho-centre,centroid and circumcentre of the triangle OPQ ?
2)What is the nature of the locus repesented by the equation (x-y+c)2+(x+y-c)2=0 ?
3)If the vertices P,Q,R of a triangle PQR are rational points,which of the following points of the triangle PQR are always rational points?(tell me wat is meant by rational points?) IIT 98
(A)Centroid (B)Incentre
(C)Circumcentre (D)Orthocentre
4)If the line y=x√3 cuts the curve x3+y3+3xy+5x2+3y2+4x+5y+1=0 at the points A,B,C then prove that the value of OA,OB,OC is equal to 4/13 (3√3-1)
-
UP 0 DOWN 0 0 19
19 Answers
1
1hey i got the method for the 4th prob! check this!
http://www.goiit.com/posts/list/analytical-geometry-straight-lines-1014240.htm#1249439
1@ teena
i got that already..And the ans is A,C,D not only centroid!!
i am waiting for the solution of 4th one!! u got that??
1since the centroid is ( x1+x2+x33 ,y1+y2+y33 )
itz always rational as the vertices of the triangle are rational points....
so the ans is centroid...
1a rational point is a point whose both coordinates r rational numbers...
1yes it is teena
as u know that cirlce's general equation is x2 +y2 + 2gx+2fy + c =0
in the abv solved equation u can notice that g = 0 and f = -c
1as centre of cirlce is ( -g , -f ) u can easily see that my solved equation gives u that thng only [4]
1for the 2nd one...
(x-y+c)2+(x+y-c)2 =0
2(x2+(y-c)2) =0
x2+y2-2cy+c2 =0
x2+y2-2cy+C =0 where C=c2
so itz a circle with the centre on y- axis....
z it????
62where is O? (origin?)
You should keep in mind that you know the value of P and Q because on the x axis, y=z=0
The 2nd equation says that x-y+c=x+y-c=0 which is a point
12) he given the ans as point (0,c) {he equated x-y+c=0 and x+y-c=0) i don no why?
4)OA,OB,OC are the lengths only where O is the origin !!!
12) i guess it represents a circle as i got solving the given equation as x2 + y2 - 2yc + c2= 0which represents a circle .
4) OA, OB, OC r the lengths u talking abt or there sum or product equals to the value u asked to prove.
1@ pritish i got it!:)
@ manmay thank u very much!!
also guys can u also solve the 2nd and 4th question???????
1One thng i want to add for (1) is that since its a right angled thriangle it's orthocentre will be midpoint of it's Hypotenuese that is PQ's midpoint.
angles can be found out using slopes and as we know one angle is already 90° it is not too much of a task so Orthocentre is also found out.[1]
11) Ax + By + C = 0 cuts x-axis means y = 0 which satisfies the given equation of line, hence
Ax + 0 + C = 0 → x = -CA
hence P is ( -CA , 0 )
and similarly Q ( 0 , -CB )
hence the coordinates
we got three vertices of ΔOPQ as O ( 0,0) , P( -CA , 0 ) , Q ( 0 , -CB )
for centroid i thnk u know , for circumcentre use property let point be X
XP = XQ = XO → XP2 = XQ2 = XO 2
for orthocentre use this If A(x1,y1) ,B(x2,y2) ,C(x3,y3)
orthocentre is (x1tanA + x2tanB + x3tanCtanA + tanB + tan C, y1tan A + y2tan B + y3tan Ctan A+ tanB + tan C )
39Incentre is equidistant from all three sides of the triangle...so when you apply distance of point from line formula, you may not get x and y rational(due to the square root). It depends. The other three points do not have such problems.
1@ pritish dude thanks for the def. ur ans is correct but i have a doubt,,, y cant the incentre be rational.. the points can be in rational form naa(basing on its formula)!!!
@nishant sir! for the first how to find the vertices of P and Q..?
and for the 2nd how did u equate both? i din get ur approach!!!
39A point (x,y) in the plane is called a rational point if both x and y are rational quantities. If x and y are integers, it is called a lattice point.
Now if the three vertices of a triangle are rational, its centroid, circumcentre and orthocentre, all derived from vertices, are rational points. Incentre however may NOT be a rational point.
Another thing to note, in 2D, an equilateral triangle which has all vertices rational does NOT exist. In space it can.