Question

find the foot of the perpendicular drawn from (3,0) upon the st.line 5x+12y-41=0​

Answer 1

Answer:

1 निम्लिखित कव्यांश के आधार पर पूछे गए प्रश्नों के उत्तर एक-दो वाक्यों में लिखे।

हम बालक है सभी ढंग के,

भेद नहीं पर जाति रंग के ।

हम सारे जग को ही समझे,

प्रिय अपना घर - द्वार।

मधुर हम बच्चो का संसार ।

Answer 2

Step-by-step explanation:

For the straight line,

5x+3y-41= 0

consider the foot as (h,k).

let that point be like ,

take x= a(say), then y = 41-5a

------------

3

so that (h,k) point will be ( a, 41-5a)

--------

3

so as you know that, the given line and the line joining the points are perpendicular.

m1m2= -1 ( m --> slope)

and let m1 =slope of line joining (3,0) and our (h,k)

m2= slope of given line.

m2= - coeff of x. = -5/12

m2= - coeff of x. = -5/12 --------------

m2= - coeff of x. = -5/12 -------------- coeff of y

m1= y2-y1

m1= y2-y1 -------- = [(41-5a) / 3 ] - 0

x2-x1. ---------------------

( a. - 3)

205-25a= 36a-108

61 a= 313

a= 313 / 61.

so h= 313 / 61.

and k= 41-5(313 /61). = 2501- 1565

---------------- ----------------

3 61(3)

k = 936/ 183

(h,k) = 313. 936

( ------- , ------- ) =( 5,5) (approx)

61. 183

Madam, hope you got the answrr.,tell me correct or not . hope I have tried something !

(note : diagram - ignore ( rough work )not to scale (only reference))

Is this from 11 th class, or 10 th class?

bcoz i have seen such in my 10 th class.