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debadutta mishra Grade: 11
        

let agiven line L1 intersect the x and y axis at P &Q respectively.Let another line L2 perpendicular L1 cut the x and y axis at R & S respectively.Show that the locus of the point of intersection of the lines PS & QR is a circle passing through the origin.

7 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

let given line is L1: y=mx+c


(P,Q)=(-c/m,0) , (0,c)


now L2 is perpendicular to line L1


slope of L2*slope of L1=-1            (m1m2=-1)


 eq of L2: y=-x/m + k                 (k is any variable)


(Q,R)=(km,0),(0,k)


eq of PS is y=mkx/c + k and.................1


eq of QR is  y=-x/km + c ...................2


let the point if intersection be (X,Y)


 after solving 1 and 2


X= (c-k)/(mk/c + 1/mk )...............3


Y= (km+1/m) /(mk/c + 1/mk)...........4


 eliminate k by solving eq 3 and 4 you will get the equation in terms of X,Y and c .......this will give the required locus which is circle passing through origin...

7 years ago
Anonymous Anon
11 Points
										
Typos above: (R,S) =(km,0),(0,k)should be(Q,R)=(km,0),(0,k)
y=-cx/km + cshould be eq of QR is  y=-x/km + c
Also, no need to find intersection of PS and QR. Simply eliminate the parameter k in the equations of PS and QR to get the locus of the intersection point as line L2 slides perpendicularly on given line L1 (L1 is a fixed “given” line). From PS and QR equations,
k = cy/(mx + c) = cx/(cm – my)
cy(cm – my) = cx(mx+c)
y(cm – my) = x(mx+c)
mx^2 + my^2 + cx – mcy = 0
x^2 + y^2 + cx/m – cy = 0, which is a circle passing through the origin.
10 months ago
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