 # 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.

11 years ago

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...

5 years ago
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.