equation of a line....perpendicular to....

[candy101]

ok so i am having big problems with this one


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

I don't know if i did the steps right but...i tried

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 really know what else to do (im not good with fractions)

 

[wjm11]

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

I don't know if i did the steps right but...i tried

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 really know what else to do (im not good with fractions)

Good work so far. you're almost done. You found the slope: 3/4. So the equation is y = (3/4)x + b. You substituted in the point (-1,3), and you are ready to solve for b.

3= 3/4 (-1) + b
3= -3/4 + b
b = 3 + 3/4
b = 3.75

So

y = .75x + 3.75 or
y = (3/4)x + 15/4 if you prefer.