Question

line 3x+y=12 intersects line x+2y=4 at P. Given points Q(0,3) and R(3,9), then equation of the line passing P perpendicular to QR is?

Answer 1

where are you stuck at ?

Answer 2

Steps :
(A)Q(0,3) and R(3,9), find the slope of line QR using slope formula.

(B)Find slope of line through P, because it is perpendicular to QR.
(product of slopes of perpendicular lines = -1)

Hence you get the slope, now you only need a point to fins the equation.

(C) Solve 3x+y=12 and x+2y=4 simultaneously. This will give you point of intersection P.
now you have both slope and point,
(D) find the equation of line using \(y-y_1 = m (x-x_1)\)

Answer 3

do you mean slope is gradient?

Answer 4

generally while doing vector calculus we use the term gradient, else they both are same.

Answer 5

emmm oky i am calculating nw!!

Answer 6

will the answer be y=3x-12 ??

Answer 7

"(A)Q(0,3) and R(3,9), find the slope of line QR using slope formula. "

i am sure you got the slope as 2.

Now, the required line is PERPENDICULAR to line QR

so the slope of the required line will be \(\Large \dfrac{-1}{2}\)

got this ?

Answer 8

kaha chali jaati hai bar bar! :O
problem to khatm karke jaati :P (i couldnt understand what are you telling sorry)
( i know whr i went wrong i gt the answer ready its 2y+x-2=0 )

Answer 9

am i right @hartnn ?

Answer 10

you mean 2y+x-4 =0
??
and i thought you knew hindi :P

Answer 11

because 2y+x-2=0 is incorrect !
what did u get as co-ordinates of P