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Write an equation in slope-intercept form of a line with the following characterisitcs; perpendicular to the graph of:
2x - 3y = 12 and passes through (4,3).

2007-07-24 04:16:40 · 6 answers · asked by Amber B 1 in Science & Mathematics Mathematics

6 answers

Take the eqn given and put in slope-intercept form:
-3y = -2x + 12
y = (2/3)x - 4
The slope of required line will have a slope equal to the negative reciprocal of the given line, call it m = -3/2
The required line is given by
y(x) = (-3/2)x + b where b is the y-intercept
To pass thru the point (4,3):
y(4) = (-3/2)(4) + b = 3
-6 + b = 3
b = 3+6 = 9
y(x) = (-3/2)x + 9 is the required line

2007-07-24 04:25:35 · answer #1 · answered by kellenraid 6 · 0 0

Re-arrange 2x - 3y = 12 to
3y = 2x + 12
y = (2/3)x + 4

line equation
y = mx+b

perpendicular to the line y = (2/3)x + 4 means
m* (2/3) = -1

so, m = - 3/2

therefore,
y = (-3/2)x + b

Passes through the point (4,3), so
y = (-3/2)x + b
3 = (-3/2)(4) + b
b = 9

y = (-3/2)x + b
y = (-3/2) x + 9

2007-07-24 04:28:36 · answer #2 · answered by buoisang 4 · 0 0

2x - 3y = 12 has a slope of 2/3
So a line perpendicular to it will have a slope of -3/2.
Thus its equation will be y = -3x/2 + b
Since it goes through point (4,3) we have:
3 = -3/2*4 = b so be = 3+6 = 9
Hence y = -3x/2 +9 is the required equation.

2007-07-24 04:27:58 · answer #3 · answered by ironduke8159 7 · 0 0

2x-3y=12---> -3y =12 -2x or y = -2/3 *x -4
the slope of that line is -2/3 The permendicular line ha a slope
a= -1/-2/3 =3/2
form of the line y =(3/2)*x +b (1)
replace in (1) y by 3 and x by 4

3 = (3/2)*4 + b or 3 =6+b so b =-3

the line y = (3/2)x -3

2007-07-24 04:27:06 · answer #4 · answered by maussy 7 · 0 1

-3y=-2x+12
y=(2/3)x-4
slope is 2/3
slope of perpendicular line is -3/2
y=(-3/2)x+b
it passes through (4,3)
3=(-3/2)(4)+b
b=9
the equation is y=-(3/2)x+9

2007-07-24 04:29:05 · answer #5 · answered by cidyah 7 · 0 0

y = 2/3 x - 4

i.e. Gradient = 2/3
Y-intercept = -4

2007-07-24 04:21:55 · answer #6 · answered by Doctor Q 6 · 0 1

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