# find the equation of the line...perpendicular to....

• Dec 12th 2009, 09:57 PM
Candy101
find the equation of the line...perpendicular to....
the question says==> find an equation of the line through (-1,3) which is perpendicular to 4x+3y=12

and i got this===> y=-4/3x+b ==> i flip the -4/3 to 3/4

then putting it all together i got ==> 3= 3/4 (-1) + b

but thats as far as i can go.. i don't reall know what else to do (im not good with fractions)

how would i multiply the fraction and add it with whole number?

(there is a thread tht is something like this already and i posted in it, bc i didn't wanted to start a new one, but i guess i can)

Thanks ,
• Dec 12th 2009, 10:39 PM
Quote:

Originally Posted by Candy101
the question says==> find an equation of the line through (-1,3) which is perpendicular to 4x+3y=12

and i got this===> y=-4/3x+b ==> i flip the -4/3 to 3/4

then putting it all together i got ==> 3= 3/4 (-1) + b

but thats as far as i can go.. i don't reall know what else to do (im not good with fractions)

how would i multiply the fraction and add it with whole number?

(there is a thread tht is something like this already and i posted in it, bc i didn't wanted to start a new one, but i guess i can)

Thanks ,

Rearrange $\displaystyle 4x+3y=12$ to $\displaystyle y=-\frac{4}{3}x+4$

m=-4/3

the gradient of the line perpendicular to this is 3/4

then we know this line passes through (-1,3)

using this formula $\displaystyle y-y_1=m(x-x_1)$

$\displaystyle y-3=frac{3}{4}(x+1)$

simplify it .