Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, - Two-dimensional Coordinate Geometry - BITSAT-2

The coordinates of the foot of perpendicular from (a,o) on the line y= mx+\frac{a}{m}   are  

  • Option 1)

    \left ( 0,\frac{-a}{m} \right )

  • Option 2)

    \left ( \frac{a}{m},0 \right )

  • Option 3)

    \left ( 0,\frac{a}{m} \right )\:

  • Option 4)

    None

 

Answers (1)

best_answer

 

General form of the equation of a line -

In  Ax+By+C=0 , scope of a line = \frac{-A}{B} .

- wherein

a,b,c  are the constants.

 

m^{2}x-my+a =0 -------------------------(i)

step   l=m

Step of 1^{r} = \frac{-1}{m}

 

so 1^{r} = equation y = \frac{-1}{m}\left ( x-a \right )

\Rightarrow x+my-a=0 ------------------------ (ii)

Solutions (i) and (ii)

\left ( m^{2} \right+1 )x =0 \Rightarrow x=0

y= \frac{a}{m}


Option 1)

\left ( 0,\frac{-a}{m} \right )

Incorrect

Option 2)

\left ( \frac{a}{m},0 \right )

Incorrect

Option 3)

\left ( 0,\frac{a}{m} \right )\:

Correct

Option 4)

None

Incorrect

Posted by

Plabita

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT iteration 1 result 2025 out on bitsadmission.com; confirm allotment by July 14
anam-khan

BITSAT 2025 preference editing window open till July 5 for BE, BPharm, MSc admission
anam-khan

BITSAT cut-off for BE Computer Science slightly drops, increases for all others; session 2 result soon

Latest Question