equation of the normal of the line on point..

**jvignacio** · September 15th 2008, 09:37 PM

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!!

**Chris L T521** · September 15th 2008, 09:43 PM

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.

--Chris

**jvignacio** · September 15th 2008, 10:13 PM

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.

--Chris

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

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