write the equation of the line in slope-intercept form that is perpendicular to the graph of 3x+y=5 and goes through the point (-9,3)

Answer 1

perpindicular lines have slopes with two characteristics 1.) opposite 2.) reciprocal fractions (flipped over)

so to find the slope on the perpindicular line we must be able to read off the slope of original equation 3x+ y =5

this involves solving for y to put equation into y = mx + b and the slope will be coefficient mult to x variable

3x + y = 5

-3x -3x

____________

y = -3x +5 so now original m = -3 so perpindicular m = 1/3 has opposite sign and fraction flipped

Now the next thing is to make this line go through the point (9,-3) there are several ways to do this

one way is to use point slope equation y - y1 = m(x -x1) now y and x are not substituted

but (9, -3) become (x1, y1)

y - (-3) = 1/3(x - 9) simplifying

y+3 = 1/3x - 9*1/3

y+3 = 1/3x -3

-3 -3

y = 1/3x - 6

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Now some students have learned this process a different way and just plug (9,-3) and slope into y= mx +b

which is certainly ok but if question had specifically asked to use point slope equation would have been incorrect or points taken off

solving for b in this manner -3 = 1/3(9) + b

-3 = 3 + b

-3 -3

___________

-6 = b

y = 1/3x -6 arrives at the same answer.

Answer 2

3x+y=5

solve y

y =-3x +5, then the coefficient of the x term is the slope

a perpendicular line has a negative inverse slope = 1/3

y = x/3 + b, plug in the point (-9,3) to solve for b

3=-9/3 +b

3=-3+b

b= 3+3 = 6

y = x/3 + 6 is the perpendicular line through the given point

or more explicitly in slope intercept form

y = (1/3)x +6 where 1/3 = m and 6=b for the general form y=mx+b