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

Well, first, solve your 2x - 3y = 12 equation for y, i.e., make it in conventional slope-intercept form. Then, once you find your slope (m), take the negative reciprocal (-1/m) and that will be the slope of the line perpendicular. If it passes through point (2,6), so you can plug those values into your new equation and solve for B (the y-intercept), and then you will be able to fully-build your new equation. Try this and tell me what you get!

Answer 2

first transform the given equation to the general format of
y = mx + c
where
m = slope
c = intercept
and u have to remember this also
when there r 2 lines which r perpendicular to each other ,
if the slope of one line is "m" then the slope of the other is " (-1/m)"
so...
moving forward..
when u transform
2x - 3y = 12 into
y = mx + c format
u will get...
3y = 2x -12
y = (2/3)x - 4

now remember what i mention about the slope..
if the slope of one is "m" then the slope of the other one is " -1/m"

according to that the slope of the line we want to draw will be
(-3/2)
then we can write the equation of the line we want as..
y= (-3/2) x + c

let's consider the given points (2 , 6)
by substituting them in the equation
we get
6 = (-3/2)2 + c
then c will be
c = 9
therefor the equation of the line will be
y = (-3/2)x + 9

Answer 3

thanks both of you for your explinations on how to solve them :)