writing a slope intercept equation

Question

write the slope intercept equation of the line that is perpendicular to 3y+x=3 and passes through (6,-1)

 

Answer 1

The equation of the line is

3y+x=3 which is y=-(1/3)x + 3

the slope of the line is (-1/3) and the one of the perpendicular will be

3

The equation then will be

Y = 3x + b

replacing x and y of the given point (6,-1) we obtain:

-1= 18 +b

b= -19

The equation then will be:

Y = 3X -19

 

 

 

 

Answer 2

a perpendicular line has a negative reciprocal slope.

First we need to determine the slope of the given line.

Write it in slope-intercept form

 

3y + x = 3

3y = -x + 3

y = -(1/3)x + 1

We see the slope is -1/3.

Therefore the slope of the perpendicular line is 3 (the reciprocal of 1/3 is 3/1 or 3 and we multiplied by -1)

 

So far we have y = 3x + b

We find the value of b, the y-intercept by plugging in the coordinates of the point the is on the line

 

-1 = 3(6) + b

-1 = 18 + b

b = -19

 

Therefore the slope-intercept equation is y = 3x - 19