How do you write an equation in slope-intercept form for an equation of a line perpendicular to y = -3x +1y=3x+1 and that intersects at point (1, 2/3)(1,23)?

1 Answer
Mar 29, 2018

3y=x+13y=x+1

Explanation:

The slope of this line is -33. Slope of a line perpendicular to it it 1/313.
Because m_1*m_2=-1m1m2=1 where m_1m1 and m_2m2 are the slope of perpendicular lines

Now, the equation of the line is y=x/3+cy=x3+c

To find cc, put a point which in this case is (1,2/3)(1,23)

2/3=1/3+c23=13+c => We get that c=1/3c=13

So required equation is y=x/3+1/3y=x3+13 or 3y=x+13y=x+1