Question

The locus of feet of perpendicular drawn from origin to the straight lines passing through (2, 1) is

A)
x2 + y2 - 5y = 0

B)
x2 + y2 - 2x - y = 0

C)
2x + y - 5 = 0

D)
(x - 2)(y - 1) = 0​

Answer 1

$\sf{\underline{\boxed{\pink{\large{\bold{ ♡\: HELLO \:♡}}}}}}$

$\green{\large\underline{\underline\mathtt{Given:}}}$ A straight line is passing through the Point (2,1).

$\blue{\large\underline{\underline\mathtt{RTF:}}}$ Locus of feet of perpendicular drawn from origin to the line.

$\sf{\underline{\boxed{\pink{\large{\bold{ ◇SOLUTION◇}}}}}}$

y=mx+c

1=mx+c

In the figure, AP and OP are perpendicular ,

☆So m1.m2 =-1

=>k/h × (k-1 /h-2) = -1

=>k (k-1) = -h (h-z)

$= > y {}^{2} - y = - x {}^{2} + 2x$

Hence,

$= > {\boxed{x {}^{2} + y {}^{2} - 2x - y = 0 }}$

HOPE IT HELPS :)

Answer 2

ʜᴇʏ ʙᴇsᴛɪᴇ☻︎!!!

Answer:

ᴜʀ ᴀɴsᴡᴇʀ ɪs ɪɴ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ !!!ᴘʟᴢ ᴍᴀʀᴋ ɪᴛ ᴀs ʙʀᴀɪɴʟɪᴇsᴛ ɪғ ɪᴛ ʜᴇʟᴘᴇᴅ ʏᴏᴜ..ᴀɴsᴡᴇʀ ɪs " ʙ"