Analytical Geometry> The locus of the foot of the perpendicula...

The locus of the foot of the perpendicular from the origin to the line which always passes through a fixed point (h,k) is

John Doe , 11 Years ago

Grade 12th pass

3 Answers

Arun Kumar

Last Activity: 11 Years ago

Hello Student,

y = mx+c
Foot of perpendicular from (0, 0) to the line be (a, b)

$a = \frac{-mc}{1+m^{2}}$

$b = \frac{c}{1+m^{2}}$

$m = \frac{-a}{b}$

Since line passes through (h, k):

$k = mh +c$

$c = k-mh$

$b = \frac{k-mh}{1+m^{2}}$

$b = \frac{k-(\frac{-a}{b})h}{1+(\frac{-a}{b})^{2}}$

$a^{2}+b^{2}-ah-bk=0$

$x^{2}+y^{2}-ax-by=0$

$a\rightarrow x, b\rightarrow y$

=>Equation of circle.

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Approved

ram

Last Activity: 8 Years ago

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p s bose

Last Activity: 8 Years ago

Equation of any line through point (h,k) is y-k=m(x-h)
Equation of the line perpendicular to the line through origin is y=(-1/m)x
To get the locus of the intersection of the above two lines i,e foot of the perpendicular to first line, eleminate ‘m’ from the above two equations.
Multiplying the two equations, y(y-k)=m(-1/m)(x-h)x which is same as x^2 + y^2 – hx – ky = 0