equation of the line that is perpendicular to -3x+2y=-6 that passes through the point (3,3).

Question

what would the equation be for this line in algebra solution.

Answer 1

-3x + 2y = -6

 

Let's put this into slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept

 

      -3x + 2y = -6

      2y = 3x - 6

 

      y = (3/2)x - 6

 

The slope is 3/2.  A line that is perpendicular will have a slope equal to the negative reciprocal (-1/m).  The negative reciprocal of 3/2 is -2/3.  So the equation of the perpendicular line so far is:

 

      y = (-2/3)x + b

 

To find b, plug in the given point, (3,3), and solve for b:

 

      3 = (-2/3)(3) + b

      3 = -2 + b

      5 = b

 

So the answer is:

 

      y = (-2/3)x + 5

Answer 2

So first you will want to put the equation into slope-intercept form (y=mx+b). 

-3x+2y=-6  Add 3x to both sides.

2y=3x-6   Then divide by 2.

y=3/2x-6  Now it is in slope-intercept form.

 

The slope for this particular equation is 3/2 (m).

The slope of a line that is perpendicular to this would be -2/3, the negative reciprocal of 3/2.

 

Next you need to plug (3,3) and the slope into the slope-intercept form to find the y-intercept (b).

3=-2/3(3)+b

3=-2+b  Add 2 to both sides.

5=b  Now you know your y-intercept.

 

Now you have your final equation: y=-2/3x+5  :)