Given:
The equation of line p is
[tex]y=-\dfrac{1}{3}x-3[/tex]
Line p and q are parallel.
To find:
The equation of line q.
Solution:
The slope intercept form of a line is
[tex]y=mx+b[/tex]
Where, m is slope and b is y-intercept.
The equation of line p is
[tex]y=-\dfrac{1}{3}x-3[/tex]
The slope of the line is [tex]-\dfrac{1}{3}[/tex].
We know that the slopes of parallel lines are equal.
Line p and q are parallel. So,
Slope of line q = [tex]-\dfrac{1}{3}[/tex]
Line q passes through (6,-4) with slope [tex]-\dfrac{1}{3}[/tex], so the equation of the line is
[tex](y-y_1)=m(x-x_1)[/tex]
Where, m is the slope.
[tex](y-(-4))=-\dfrac{1}{3}(x-6)[/tex]
[tex]y+4=-\dfrac{1}{3}x+2[/tex]
[tex]y=-\dfrac{1}{3}x+2-4[/tex]
[tex]y=-\dfrac{1}{3}x-2[/tex]
Therefore, the equation of line q is [tex]y=-\dfrac{1}{3}x-2[/tex].
