2007-08-27 23:19:53 · 4 answers · asked by Bhaskar 4 in Science & Mathematics ➔ Mathematics
normal is perpendicular to the point.
thus the gradient would be -1/gradient of perpendicular line = 1/4.
y=mx+c
y=x/4 + c
to find c, sub the coord. of the points given.
so... -3=2/4 +c
-7/2=C
thus the eqn should be :
y=x/4 - 7/2
to simplify:
y=x/4 - 14/4
4y=x-14 #
hope I'm correct.
bb. (:
2007-08-27 23:48:26 · answer #1 · answered by bburnouts 3 · 1⤊ 1⤋
Slope of y = 4x+ 5 is 4.
Slope of the normal is -1/4.
Therefore, equation of the normal
y - 3 = (-1/4) (x-2)
y-3 = (-1/4)x + (1/2)
4y - 12 = -x + 2
x+ 4y - 14 = 0
2007-08-27 23:51:36 · answer #2 · answered by gab BB 6 · 1⤊ 0⤋
Gradient (slope) of normal = 1 / 4
Equation of normal is:-
y + 3 = (1 / 4) (x - 2)
y = (1 / 4) x - 1 / 2 - 3
y = (1 / 4) x - 7 / 2
If desired, this may be written as:-
4y = x - 14
x - 4y - 14 = 0
2007-08-28 03:27:34 · answer #3 · answered by Como 7 · 1⤊ 1⤋
y=9x=8x+15 answer
2007-08-27 23:31:00 · answer #4 · answered by christian t 1 · 0⤊ 3⤋