find the equation of a line passing through (4,-3) which is perpendicular to 2y+3x = 4.

also, find an equation of the circle with center (-2,3) which is tangent to y=-3.

please explain how to do them also.. :)

2y + 3x = 4

Solve for y because we want to know the slope.

y = 4/2 - 3/2x

The slope of the line is -3/2.

If the slope of some line is a/b, then the perpendicular (or normal) line is -b/a.
So, the slope of the normal line will be 2/3.

y = 2/3*x + c where c is some unknown constant

The known point is (4,-3).
y = -3 and x = 4, so plug them in

-3 = 2/3*4 + c
-9 = 2*4 + 3c
c = -17/3

Therefore, y = 2/3*c - 17/3

Oh yes... Lambo.. you remember how I did that second problem. Just do it that way... way unobvious and without calculus, which would have been more obvious than my way :p

Originally posted by 4DHyperCubix
Oh yes... Lambo.. you remember how I did that second problem. Just do it that way... way unobvious and without calculus, which would have been more obvious than my way :p

thank you mr james sibley

"The circle with center (-2, 3) tangent to the line y= -3" is easy because the line y= -3 is horizontal. The line from (-2, 3) perpendicular to y= -3 is vertical: it is x= -2 and crosses y= -3 at
(-2, -3). The distance from (-2,3) to (-2, -3) is 3-(-3)= 6 so we are seeking the equation of a circle with center at (-2, 3) and radius
6. That equation is (x+2)^2+ (y-3)^2= 36.

Son of a bitch.. that is why I was missing the obviousl solution Lambo... I kept reading it as one point it was tangent at was y = -3.

Shit...

However, my other way is neat :)

Lol, stop doing his homework:)