please explain this step regarding this question- A line passes through s fixed point (a,b). The locus of the foot of the perpendicular on it from origin
y = mx+c
Foot of perpendicular from (0, 0) to the line be (a, b)

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

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

Question

please explain this step regarding this question- A line passes through s fixed point (a,b). The locus of the foot of the perpendicular on it from origin

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

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

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

Vikas TU · Answer

Dear Student,
line passing through (a,b) is y= mx-am+b,
line normal to this and passing through 0,0:
y=-x/m
let foot of perpendicular be (h,k)
we know: k= mh-am+b and
m= -h/k,
=>k= (-h/k)*h + a(h/k) +b,
so=k2=-h2+ah+bk,
So h2+k2= ah+bk,
therefore locus is x2+y2=ax+by [ans]

Cheers!!

Regards,

Vikas (B. Tech. 4^th year
Thapar University)

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please explain this step regarding this question- A line passes through s fixed point (a,b). The locus of the foot of the perpendicular on it from origin y = mx+c Foot of perpendicular from (0, 0) to the line be (a, b)

please explain this step regarding this question- A line passes through s fixed point (a,b). The locus of the foot of the perpendicular on it from origin
y = mx+c
Foot of perpendicular from (0, 0) to the line be (a, b)

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

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

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please explain this step regarding this question- A line passes through s fixed point (a,b). The locus of the foot of the perpendicular on it from origin y = mx+c Foot of perpendicular from (0, 0) to the line be (a, b)

please explain this step regarding this question- A line passes through s fixed point (a,b). The locus of the foot of the perpendicular on it from originy = mx+cFoot of perpendicular from (0, 0) to the line be (a, b)

1 Answers

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please explain this step regarding this question- A line passes through s fixed point (a,b). The locus of the foot of the perpendicular on it from origin
y = mx+c
Foot of perpendicular from (0, 0) to the line be (a, b)

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

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