Question

find the equation of the line that is perpendicular to 3x+5y=9 and passes through (2,3).

Answer 1

3x + 5y = 9

5y = -3x + 9

y = (-3x/5) + (9/5)

m = -3/5, perpendicular m = 5/3

m = 5/3 pt = (2, 3)

y = mx + b

3 = (5/3)(2) + b

3 = (10/3) + b

3 - (10/3) = b

(9 - 10)/3 = b

-1/3 = b

y = (5x/3) - (1/3)

or

3y = 5x - 1

or

-5x + 3y = -1

 

5x - 3y = 1

Answer 2

through(1,6),perpendicular to 3x+5=1

Answer 3

I suppose you mean 3x + 5y = 1

Rearranging, we have:

5y = -3x + 1
y = (-3/5)x + (1/5)

The slope of this line is thus -3/5.

The slope of a line perpendicular to this line will be -1 / (-3/5) = 5 / 3

Using the slope formula, the line which we need will be:

(y - 6) / (x - 1) = 5 / 3
y - 6 = (5/3)(x - 1)
y - 6 = (5/3)x - (5/3)
y = (5/3)x + 6 - (5/3)
y = (5/3)x + (13/3)

Answer 4

write an equation perpendicular to -4y-3x=8

Answer 5

write the equation of a line that passes through the point (2, 3) and is parallel to the line 12x-5y=2