# Finding a line perpendicular to another line

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• Dec 12th 2009, 05:08 PM
Candy101
Finding a line perpendicular to another line
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?
• Dec 12th 2009, 07:44 PM
Stroodle
You can use the formula $\displaystyle y-y_1=m(x-x_1)$

So if you sub in the point $\displaystyle (-1,3)$ and use $\displaystyle m=\frac{3}{4}$ you get:

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

$\displaystyle y=\frac{3}{4}x+\frac{15}{4}$
• Dec 12th 2009, 07:57 PM
Candy101
Quote:

Originally Posted by Stroodle
You can use the formula $\displaystyle y-y_1=m(x-x_1)$

So if you sub in the point $\displaystyle (-1,3)$ and use $\displaystyle m=\frac{3}{4}$ you get:

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

$\displaystyle y=\frac{3}{4}x+\frac{15}{4}$

how did you get 15/4?
• Dec 12th 2009, 09:13 PM
Stroodle
$\displaystyle y-3=\frac{3}{4}(x+1)$

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

$\displaystyle y=\frac{3}{4}x+\frac{3}{4}+3$

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

$\displaystyle y=\frac{3}{4}x+\frac{15}{4}$