# equation of the normal of the line on point..

• Sep 15th 2008, 09:37 PM
jvignacio
equation of the normal of the line on point..
find the equation of the line (3, -5) and (-1,2)

i did it and got an equation y = -7/4x + 1/4

how its asking to find the equation of the normal of this line at point (3, -5)

how do i get that?

i know u use m1 and m2, u have y = m2x + b.. how do u work out b?

m2 = 4/7 and x = 3 but b ?

thank u!!
• Sep 15th 2008, 09:43 PM
Chris L T521
Quote:

Originally Posted by jvignacio
find the equation of the line (3, -5) and (-1,2)

i did it and got an equation y = -7/4x + 1/4

how its asking to find the equation of the normal of this line at point (3, -5)

how do i get that?

i know u use m1 and m2, u have y = m2x + b.. how do u work out b?

m2 = 4/7 and x = 3 but b ?

thank u!!

Since the equation of the normal line is $y=\tfrac{4}{7}x+b$, to find b, plug in the point $(3,-5)$.

This tells us that $-5=\tfrac{4}{7}(3)+b\implies b=\dots$

Take it from here. :D

--Chris
• Sep 15th 2008, 10:13 PM
jvignacio
Quote:

Originally Posted by Chris L T521
Since the equation of the normal line is $y=\tfrac{4}{7}x+b$, to find b, plug in the point $(3,-5)$.

This tells us that $-5=\tfrac{4}{7}(3)+b\implies b=\dots$

Take it from here. :D

--Chris

ohh damnit lol why didnt i think of that. make b the subject