Question

Write an equation of a line in slope intercept form that is perpendicular to the line 2x -3y = 12 and passes through the point (2, 6).

Answer 1

The equation of the original line is y = 2x-4;
this equation is in the form y = mx+c,
where m is the slope of the line.
As a result, the slope of the line is 2.

The question requires the finding of the equation of the line that is perpendicular to the original line; as a result, we have to find the slope of the perpendicular line first.

Since

the slope of a line * the slope of its perpendicular line = -1;

therefore,

2 * the slope of the perpendicular line = -1
the slope of the perpendicular line = -1/2.

Given the perpendicular line passes through the point (-6,2),
by using point slope form, which is

y-y1 = m(x-x1)

By substituting the point (-6,2) into the point slope form,

y-2 = (-1/2)*[x-(-6)]
y-2 = (-1/2)*(x+6)
y-2 = (-1/2)(x)-3
y = (-1/2)(x)-1;

Therefore,
the slope-intercept form of the perpendicular line is

y = (-1/2)(x)-1.
source: http://answers.yahoo.com/question/index?qid=20081114094710AAQ5XEz

Answer 2

\[2x-3y=12\]\[2x-12=3y\]\[\frac{2x-12}{3}=y\]\[\frac{2x}{3}-\frac{12}{3}=y\]\[\frac{2x}{3}-4=y\]

Answer 3

In this equation, the slope is \[\frac{2}{3}\]
The slope of the function that would be perpendicular to this function would be\[\
-\frac{3}{2}\]

Answer 4

Thank you all(: