find the equation of the line that is perpendicular to the given line & passes thru the point(-6,5), 2x-3y=27?

Answer 1

(-6,5), 2x-3y=2

Let’s get this equation in Slope-Intercept form…
2x-3y=2

Add -2x to both sides
-3y = 2 + -2x

Multiply both sides by -1/3
y = 2x/3 –2/3

Now we have the slope of the given line, 2/3.
A line perpendicular to that line will have a slope that is the negative reciprocal of this one… that is, -3/2

So, our perpendicular line will be
y = -3x/2 + b

We have a point on the line, (-6, 5)
So, plugging in these values for x and y into onto the equation for the perpendicular line,
5 = -3(-6)/2 + b
5 = 9 + b
b = 5 + -9 = -4

So, the equation for the perpendicular line is…
y = -3x/2 - 4

Answer 2

Solve the equation for y:
-3y = -2x + 27
y = (2/3)x - 9
Slope of given line is 2/3 & y-int is -9.
Slope of line perpendicular to given line is negative reciprocal: -3/2
So new equation is:
y = (-3/2)x + b
To solve for y-int, substitute (-6,5) in for (x,y) in new equation and solve for b:
5 = (-3/2)(-6) + b
5 = 9 + b
-4 = b
New equation is y = (-3/2)x - 4.
If it has to be in the same form as the given equation, you can multiply through by 2 to get
2y = -3x - 8, then add 3x to each side to get
3x + 2y = -8.

Answer 3

-3y = - 2x + 27
y = (2/3).x - 9
Gradient of perpendicular line = (- 3/2)
Equation of perpendicular thro` (- 6,5) is:-
y - 5 = (-3/2).(x + 6)
y = (-3/2).x - 4
or
(3/2).x + y = (- 4)
3x + 2y = - 8